If $x-y$ were written as $-y+x$, then would its conjugate be $-y-x$? The conjugate of $x-y$ is $x+y$. But, if I wrote it as $-y +x$, would the conjugate be $-y-x$?
Furthermore, is the conjugate of $x + y +z$

*

*$(x+y)+z \to (x+y) - z $


*$(x+z)+y \to (x+z) - y $


*$(z+y)+z \to (z+y) - x $
Is there a way to formalize the idea of a conjugate?
 A: You can indeed formalize the notion of "conjugate", in several ways. But it's not "change the sign of the second term".
If you have a complex number $z = x+iy$ then its conjugate is $x-iy$. If you write $z = iy + x$ then the conjugate is still
$$
\bar z = x -iy = -iy +x .
$$
When you are dealing with radical expressions like
$a + b\sqrt{2}$ you find the conjugate by replacing the square root by its negative. That's a different conjugate, also often useful.
A: I think the confusion arises because you don't have a definition of conjugate. Unfortunately, a "cool enough" definition of conjugate is quite abstract. I think it's still a good idea to try to get a feeling of the argument behind conjugation without understanding all the details.
If you know some field theory, the field-theoretic definition of "conjugate" goes as follows: given a Galois field extension $K/F$ and an element $\alpha \in K/F$, you can look at the Galois group $\mathrm{Gal}(F(\alpha)/F)$, and for any $\sigma \in \mathrm{Gal}(F(\alpha)/F)$, the operation $\alpha \mapsto \sigma(\alpha)$ will be called a "conjugation map" and the element $\sigma(\alpha)$ is called a conjugate of $\alpha$. When $[F(\alpha):F] = 2$ (or in other words, when the extension is quadratic and Galois), the group $\mathrm{Gal}(F(\alpha)/F)$ has order $2$, and its non-trivial element $\sigma$ sends $\alpha$ to "its (non-trivial) conjugate" $\sigma(\alpha)$.
If this made no sense to you, just know that in the most famous quadratic extensions that you might have heard of, such as $\mathbb C / \mathbb R$ or $\mathbb Q(\sqrt 2)/\mathbb Q$, the conjugation operation does what you think it does, namely
$$
a+bi \mapsto a-bi, \qquad a+b \sqrt 2 \mapsto a-b\sqrt 2.
$$
What it does behind the scenes is that it looks at the roots of the minimal polynomial of the element and "swaps the roots". So for instance, the polynomial with real coefficients of minimal degree that has $a+bi$ as a root (where $a,b \in \mathbb R$) is
$$
(x-(a+bi))(x-(a-bi)) = x^2 - 2ax + (a^2+b^2) = 0
$$
and the conjugation operation maps one root of that polynomial to the other one. The same idea happens for $\mathbb Q(\sqrt 2)/\mathbb Q$; the polynomial with rational coefficients of minimal degree that has $a+b\sqrt 2$ as a root (where $a,b \in \mathbb Q$) is
$$
(x-(a+b\sqrt 2))(x-(a-b\sqrt 2)) = x^2 - 2ax + (a^2-b^2) = 0
$$
and again, the conjugation operation maps one root to the other.
The first moment calculus students usually encounter the word "conjugate" is when trying to work with difference of square roots, something like
$$
\lim_{x \to \infty} \frac{\sqrt{2x+1} - \sqrt{2x-1}}2 
= \lim_{x \to \infty} \frac{(\sqrt{2x+1} - \sqrt{2x-1})(\sqrt{2x+1} + \sqrt{2x-1})}{2(\sqrt{2x+1} + \sqrt{2x-1})} \\
= \lim_{x \to \infty} \frac{(2x+1)-(2x-1)}{2(\sqrt{2x+1} + \sqrt{2x-1})}
\underset{x \to \infty}{\longrightarrow} 0.
$$
This is a general trick, and I would argue that the use of the word "conjugate" is interpreted by most people as "the technique I use to clear the square roots", but it has nothing to do with the change of sign $(x+y) \mapsto (x-y)$ and more to do with the identity $x^2-y^2 = (x-y)(x+y)$ where you want to consider "the other factor" and make it appear in your problem to square both $x$ and $y$, which are usually wrapped in a square root in some way ($\sqrt 2$, $i = \sqrt{-1}$, etc.). The other factor is called "the conjugate factor".
Formally speaking though, this is always at least the image of some element considered a conjugate of something in some field extension. Let $F$ be a field, and consider the field extension $F(\sqrt b)/F(b)$ where $b$ can be anything (an element of $F$, an algebraic element over $F$, transcendental, indeterminate, whatever). If $\sqrt b$ is not in $F(b)$ (or in other words, if you can't compute the square root within $F(b)$), this extension is Galois of degree $2$ with minimal polynomial $m_{\sqrt b}(t) = t^2 - b$. Over $F(\sqrt b)$, this polynomial splits as $t^2-b = (t-\sqrt b)(t+\sqrt b)$, so that $-\sqrt b$ is conjugate to $\sqrt b$. Any element of $F(\sqrt b)$ can be uniquely written as $u+v \sqrt b$ where $u,v \in F(b)$. The conjugation operation is what you expect it to be, namely $u+v \sqrt b \mapsto u - v \sqrt b$. This is a field automorphism of $F(\sqrt b)$ which fixes $F$, i.e. an element of $F(\sqrt b)$ is equal to its own conjugate if and only if $v=0$, i.e. if and only if the element belongs to the fixed field $F(b)$ of the automorphism. In the field extension $F(t,\sqrt b)/F(t,b)$, elements can be uniquely written as $u(t) + v(t) \sqrt b$ where $u(t), v(t)$ are ratios of polynomials in $t$. The conjugation operation still exists here (for the same reasons) and it maps $(t+\sqrt b)$ to $(t-\sqrt b)$ and vice-versa.
All this to say that even in the most abstract settings, the equation $(x^2-y^2) = (x-y)(x+y)$ is still of crucial importance, and the two factors $x-y$ and $x+y$ are called conjugate of each other. If you should remember one thing of that long discussion, it is that conjugation has more to do with $x^2-y^2$ than it has to do with changing the sign of $x+y$ to $x-y$.
Évariste Galois himself studied all the possible ways one could look at a general multivariate polynomial (not just $x^2-y^2$) and see if there were any restrictions on how to permute the factors while respecting the algebraic structure that the elements $x$ and $y$ could have. Turns out there are, because if there were no restrictions, the number of permutations would always be $n!$ (such as $2! = 2$ in the case of $x^2-y^2 = (x-y)(x+y) = (x+y)(x-y)$). Some Galois extensions are very different from this, and it's not easy to picture them without diving into the theory! When the factorization of the polynomial admits enough "good" permutations to catch all the possible permutations of the factors, the field generated by using the factors is called a Galois extension, and Galois theory (the theory of Galois field extensions) is probably one of the most influential field of abstract algebra to this day.
Hope that helps,
A: The term "conjugate" is meant to be used with numbers, where in some context there is a natural other number that they pair with. Usually, it's pairs of numbers that are roots of the same quadratic polynomial with coefficients in a more restrictive field.
So when $a,b$ are real, $a+bi$ and $a-bi$ are conjugates because they are both roots of $x^2-2ax+(a^2+b^2)$, a quadratic polynomial with real coefficients. It's not really about changing a plus sign to a minus sign.
Similarly when $a,b$ are rational, $a+b\sqrt{2}$ and $a-b\sqrt{2}$ are conjugates because they are both roots of $x^2-2ax+(a^2-2b^2)$, a quadratic with rational coefficients. Again, it's not really about changing a plus sign to a minus sign.
So if you rearrange to $bi+a$, it still has the same conjugate: $a-bi$ (or if you want to write it as $-bi+a$, fine.) Because $bi+a$ and $a-bi$ are the two roots of the same polynomial.
And with abstract $x$ and $y$ without more context, there is not a good meaning for the conjugate of $x+y$.
