Is my definition of a triangle center function "equivalent" to the usual definition? There's a definition of triangle center function already in existence, but I don't really understand it. Anyway, here's my attempt at defining this concept using ideas I'm more comfortable with.

Let $E$ denote a Euclidean space; wlog we can assume $E=\mathbb{R}^n$.
By a triangle center function on $E$, let us mean a mapping $$c : E \leftarrow E \times E \times E$$ satisfying the following axioms.
Interiality. $c(x,y,z) \in \mathrm{convexHull}\{x,y,z\}$
Commutativity. $c(x,y,z) = c(y,x,z) = \cdots$
Similarity. For all bijective similarity transforms $f : E \leftarrow E$, we have $$f(c(x,y,z)) = c(fx,fy,fz).$$

What I'd like to know is:

Question. Can anyone tell whether or not these definitions are somehow equivalent, and if so, in what sense?

 A: Your first condition is not satisfied by some triangle centers: for example, the circumcenter of a triangle can lie outside it. 
Wikipedia's definition might be a bit more understandable. 
A: Types of Triangle Center Functions
A triangle center function, as defined on MathWorld, is a function $f_{1}:\mathbb{R}^{+}\times\mathbb{R}^{+}\times\mathbb{R}^{+}\to\mathbb{R}$. It must satisfy bisymmetry in the second and third arguments: $f_{1}\left(a,c,b\right)=f_{1}\left(a,b,c\right)$ and homogeneity: there is a number $m$ such that $f_{1}\left(ta,tb,tc\right)=t^{m}f_{1}\left(a,b,c\right)$. The intent is for $a,b,c$ to represent the side lengths of a triangle. By the law of cosines, we can permit a formula for $f_{1}$ to contain some trigonometric expressions such as $\cos A$, since $\cos A=\dfrac{-a^{2}+b^{2}+c^{2}}{2bc}$ is homogeneous of degree $0$ and bisymmetric in $b$ and $c$.
MathWorld also mentions a third property, "cyclicity", but that is not actually a property of $f_{1}$. Rather, given an $f_{1}$, there is a related function, which MathWorld also seems to call a "triangle center function", which outputs a triple of numbers (separated here by colons to emphasize their homogeneity): $\mathbf{f}_{2}\left(a,b,c\right)=f_{1}\left(a,b,c\right):f_{1}\left(b,c,a\right):f_{1}\left(c,a,b\right)$. Given a triangle with side lengths $a,b,c$, the triple $\mathbf{f}_{2}$ is intended to give the trilinear coordinates of a special point.
To avoid homogeneous coordinates (such as trilinear or barycentric), and to make a definition easier to understand, we can define the following related function on points in a Euclidean space: $\mathbf{f}_{3}:E\times E\times E\to E$
by $\mathbf{f}_{3}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=$
"the point in the plane containing $\mathbf{A},\mathbf{B},\mathbf{C}$
whose trilinear coordinates with respect to $\triangle\mathbf{ABC}$ are given by $\mathbf{f}_{2}\left(\left\|\mathbf{C}-\mathbf{B}\right\|,\left\|\mathbf{A}-\mathbf{C}\right\|,\left\|\mathbf{B}-\mathbf{A}\right\|\right)$". Since this comes from $f_1$ in the standard way, I will call it "the Euclidean triangle center function associated to $f_1$ (in $E$)".
In the question, goblin defines another sort of function. As Qiaochu Yuan noted in their answer, "Interiality" (using the convex hull) is not a great axiom because of centers like the circumcenter. However, being charitable, we can make the minor change from "convex hull" to "affine hull" (a.k.a. "affine span") and consider the same question. $\mathbf{f}:E\times E\times E\to E$ is said to be a goblin center function if it's "commutative" (the function is invariant under all permutations of its arguments), "planar" ($\mathbf{f}$ always lies in the affine span of its arguments), and "similar" ($\sigma\left(\mathbf{f}(\mathbf{A},\mathbf{B},\mathbf{C})\right)=\mathbf{f}\left(\sigma(\mathbf A),\sigma(\mathbf B),\sigma(\mathbf C)\right)$ for all similarity transformations $\sigma$).

Theorem

Every Euclidean triangle center function is a goblin center function and vice versa.


Proof of first direction
For convenience, define the following implicit functions of $\mathbf{A},\mathbf{B},\mathbf{C}$: $a=\left\Vert \mathbf{C}-\mathbf{B}\right\Vert$, $b=\left\Vert \mathbf{A}-\mathbf{C}\right\Vert$, $c=\left\Vert \mathbf{B}-\mathbf{A}\right\Vert$. We can use facts about trilinear coordinates to write out $\mathbf{f}_{3}$ explicitly in terms of $f_{1}$. After choosing an origin (imagining $E$ as $\mathbb{R}^n$), we have: $$\mathbf{f}_{3}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=\dfrac{af_{1}\left(a,b,c\right)}{\lambda}\mathbf{A}+\dfrac{bf_{1}\left(b,c,a\right)}{\lambda}\mathbf{B}+\dfrac{cf_{1}\left(c,a,b\right)}{\lambda}\mathbf{C}$$
where $\lambda=af_{1}\left(a,b,c\right)+bf_{1}\left(b,c,a\right)+cf_{1}\left(c,a,b\right)$ (assuming $\lambda\ne0$).
By definition, $\mathbf{f}_3$ is "planar". Next, we prove "commutativity".
Note that $a\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)=a\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$, $b\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)=c\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$, and $c\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)=b\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$. Using $a,b,c$ to denote their values with arguments $\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$, we have $\lambda\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)=af_{1}\left(a,c,b\right)+cf_{1}\left(c,b,a\right)+bf_{1}\left(b,a,c\right)$, which equals $\lambda\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$ after applying the bisymmetry of $f_{1}$. Similarly, $$\mathbf{f}_{3}\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)=\dfrac{af_{1}\left(a,c,b\right)}{\lambda}\mathbf{A}+\dfrac{cf_{1}\left(c,b,a\right)}{\lambda}\mathbf{C}+\dfrac{bf_{1}\left(b,a,c\right)}{\lambda}\mathbf{B}=\mathbf{f}_{3}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right).$$
In order to show that $\mathbf{f}_3$ is "commutative", it remains to show that $\mathbf{f}_{3}$ remains unchanged upon a cyclic permutation of $\mathbf{A},\mathbf{B},\mathbf{C}$ (since a transposition and a cyclic permutation generate $S_{3}$). Note that $a\left(\mathbf{B},\mathbf{C},\mathbf{A}\right)=b\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$, $b\left(\mathbf{B},\mathbf{C},\mathbf{A}\right)=c\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$, and $c\left(\mathbf{B},\mathbf{C},\mathbf{A}\right)=a\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$. Thus, $$\lambda\left(\mathbf{B},\mathbf{C},\mathbf{A}\right)=bf_{1}\left(b,c,a\right)+cf_{1}\left(c,a,b\right)+af_{1}\left(a,b,c\right)=\lambda\left(\mathbf{A},\mathbf{B},\mathbf{C}\right).$$ Hence, $$\mathbf{f}_{3}\left(\mathbf{B},\mathbf{C},\mathbf{A}\right)=\dfrac{bf_{1}\left(b,c,a\right)}{\lambda}\mathbf{B}+\dfrac{cf_{1}\left(c,a,b\right)}{\lambda}\mathbf{C}+\dfrac{af_{1}\left(a,b,c\right)}{\lambda}\mathbf{A}=\mathbf{f}_{3}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right).$$
Finally, we show "similarity": let $\sigma$ be a similarity which multiplies all distances by $r$. Then $\lambda\left(\sigma\mathbf{A},\sigma\mathbf{B},\sigma\mathbf{C}\right)=raf_{1}\left(ra,rb,rc\right)+rbf_{1}\left(rb,rc,ra\right)+rcf_{1}\left(rc,ra,rb\right)=r^{m+1}\lambda\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$ by homogeneity of $f_{1}$. Hence, $$\mathbf{f}_{3}\left(\sigma\mathbf{A},\sigma\mathbf{B},\sigma\mathbf{C}\right)$$
$$=\dfrac{raf_{1}\left(ra,rb,rc\right)}{r^{m+1}\lambda}\sigma\mathbf{A}+\dfrac{rbf_{1}\left(rb,rc,ra\right)}{r^{m+1}\lambda}\sigma\mathbf{B}+\dfrac{rcf_{1}\left(rc,ra,rb\right)}{r^{m+1}\lambda}\sigma\mathbf{C}$$
$$=\dfrac{af_{1}\left(a,b,c\right)}{\lambda}\sigma\mathbf{A}+\dfrac{bf_{1}\left(b,c,a\right)}{\lambda}\sigma\mathbf{B}+\dfrac{cf_{1}\left(c,a,b\right)}{\lambda}\sigma\mathbf{C}.$$ Since the coefficients of $\sigma\mathbf{A}$,$\sigma\mathbf{B}$,$\sigma\mathbf{C}$ sum to 1 (by definition of $\lambda$), we have an affine combination, which commutes with affine transformations. Since similarity transformations on $\mathbb{R}^{n}$ are special cases of affine transformations, we have $\mathbf{f}_{3}\left(\sigma\mathbf{A},\sigma\mathbf{B},\sigma\mathbf{C}\right)=\sigma\mathbf{f}_{3}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$. Note that this does not necessarily hold when $\sigma$ is affine but not a similarity (such as a shear), as we wouldn't have been able to apply the homogeneity of $f_{1}$ in that case.

Proof of second direction
Let $\mathbf{f}$ be a goblin center function. Since $\mathbf{f}$ is affine and every point in the plane is in the affine span of $\mathbf{A},\mathbf{B},\mathbf{C}$ when they are in general position, there are real-valued functions $f_{A},f_{B},f_{C}$ always summing to $1$ such that $\mathbf{f}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{A}+f_{B}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{B}+f_{C}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{C}$. Since $\mathbf{f}$ is "commutative", we have $\mathbf{f}\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)=\mathbf{f}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$, so that $$f_{A}\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)\mathbf{A}+f_{B}\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)\mathbf{C}+f_{C}\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)\mathbf{B}$$ $$=f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{A}+f_{B}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{B}+f_{C}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{C}$$.
Since the coefficients on both sides sum to 1, we must have corresponding coefficients equal, so that $f_{A}\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)=f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$ ($f_{A}$ is bisymmetric in the second and third arguments), $f_{B}\left(\mathbf{A},\mathbf{C},\mathbf{B}\right)=f_{C}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$ ($f_{B}$ becomes $f_{C}$ when the second and third arguments are swapped). By transposing other pairs, we find several other analogous relations. In conclusion, $f_{A}$ is bisymmetric, and $f_{B}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=f_{A}\left(\mathbf{B},\mathbf{A},\mathbf{C}\right)$, $f_{C}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=f_{A}\left(\mathbf{C},\mathbf{B},\mathbf{A}\right)$. Thus, we may write $\mathbf{f}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{A}+f_{A}\left(\mathbf{B},\mathbf{A},\mathbf{C}\right)\mathbf{B}+f_{A}\left(\mathbf{C},\mathbf{B},\mathbf{A}\right)\mathbf{C}$. Since $f_{A}$ is bisymmetric, we may also rewrite this in the more elegant form $\mathbf{f}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{A}+f_{A}\left(\mathbf{B},\mathbf{C},\mathbf{A}\right)\mathbf{B}+f_{A}\left(\mathbf{C},\mathbf{A},\mathbf{B}\right)\mathbf{C}$.
Let $\sigma$ be a similarity transformation. We should have $\sigma\mathbf{f}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=\mathbf{f}\left(\sigma\mathbf{A},\sigma\mathbf{B},\sigma\mathbf{C}\right)$. Since $\mathbf{f}$, when written in terms of $f_{A}$, is an affine combination, we have $\sigma\mathbf{f}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\sigma\mathbf{A}+f_{A}\left(\mathbf{B},\mathbf{C},\mathbf{A}\right)\sigma\mathbf{B}+f_{A}\left(\mathbf{C},\mathbf{A},\mathbf{B}\right)\sigma\mathbf{C}$. Setting this equal to $\mathbf{f}\left(\sigma\mathbf{A},\sigma\mathbf{B},\sigma\mathbf{C}\right)$ and equating coefficients, we have $f_{A}\left(\sigma\mathbf{A},\sigma\mathbf{B},\sigma\mathbf{C}\right)=f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)$ for all similarity transformations $\sigma$.
Intuitively, similarity transformations should be able to send a triangle to any similar triangle, so that $f_{A}$ may only depend on the length ratios $\dfrac{b}{a}=\dfrac{\left\Vert \mathbf{A}-\mathbf{C}\right\Vert }{\left\Vert \mathbf{B}-\mathbf{C}\right\Vert }$,$\dfrac{c}{a}$,$\dfrac{b}{c}=\dfrac{\frac{b}{a}}{\frac{c}{a}}$. But a function of $\dfrac{b}{a},\dfrac{c}{a}$ is equal to a homogeneous function of degree zero of $a,b,c$: we may write $f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=\widetilde{f}\left(a,b,c\right)$ where $\widetilde{f}$ is homogeneous of degree zero.
[A bit more rigorously than the above: Once an origin is fixed, a similarity transformation has the form $\sigma\left(\mathbf{x}\right)=rM\mathbf{x}+\mathbf{t}$ where $M$ is orthogonal. After a translation, we may assume $\mathbf{A}$ is at the origin. After a potential reflection, we can assume that $\mathbf{B}-\mathbf{A}$, $\mathbf{C}-\mathbf{A}$ are a positively oriented basis for the plane they span. A rotation lets us choose the angle for $\mathbf{B}$, and then knowing the lengths tells us the angle for $\mathbf{C}$ by the law of cosines and the known orientation. Thus, $f_{A}$ can depend only on the lengths. However, dilations mean it can only depend on ratios of lengths (or immediately that it's a homogeneous-of-degree-zero function of the lengths).]
Note that $\widetilde{f}$ is bisymmetric since $f_{A}$ is. Finally, define $f\left(a,b,c\right)=\dfrac{1}{a}\widetilde{f}\left(a,b,c\right)$, which is still bisymmetric, and is homogeneous of degree $-1$ (and hence is a "triangle center function" in the sense of $f_{1}$). We have $$\mathbf{f}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)=f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{A}+f_{B}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{B}+f_{C}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{C}$$
$$=f_{A}\left(\mathbf{A},\mathbf{B},\mathbf{C}\right)\mathbf{A}+f_{A}\left(\mathbf{B},\mathbf{C},\mathbf{A}\right)\mathbf{B}+f_{A}\left(\mathbf{C},\mathbf{A},\mathbf{B}\right)\mathbf{C}$$
$$=af\left(a,b,c\right)\mathbf{A}+bf\left(b,c,a\right)\mathbf{B}+cf\left(c,a,b\right)\mathbf{C}$$
Since the coefficients of $\mathbf{A},\mathbf{B},\mathbf{C}$ sum to $1$ (by definition of $f_{A},f_{B},f_{C}$), we have $\lambda=af\left(a,b,c\right)+bf\left(b,c,a\right)+cf\left(c,a,b\right)\equiv1$, and so this says exactly that $\mathbf{f}$ is the Euclidean triangle center function associated to $f$.
