Is a Linear Transformation a Vector Space Homomorphism? I see the terms linear transformation and (vector space) homomorphism used more or less interchangeably, and the set (space) of linear transformations from V to W referred to as Hom(V, W) or equivalently as  L(V, W).
From category theory, a homomorphism is a mapping between objects that preserves structure, so a vector space homomorphism is a linear transformation as this is necessary to preserve the structure of vector addition and scalar multiplication.
But to say a linear transformation is a homomorphism seems to be an invalid assumption. For example, with real / complex inner product spaces only orthogonal / unitary linear transformations preserve the inner product structure and therefore deserve to be considered as homomorphisms.
So, should linear transformations always be referred to as such (not homomorphisms), and the set of linear transformations from V to W as  L(V, W), not Hom(V, W) ?
What then about a vector space isomorphism ?
 A: They are exactly homomorphisms of vector spaces (preserving linear structures). $\mathbb{R}^n$ and $\mathbb{C}^n$ with inner products  are not plain vector spaces, they are Euclidean/Hilbert spaces. So their homomorphism would need to preserve inner products also (hence orthogonal/unitary).
Regarding isomorphisms, there is a theorem that $f:V\to W$ is an isomorphism if and only if $\dim V=\dim W$. And in that view, we often study isomorphisms of the form $f:V\to V$. These are denoted by $\mathrm{Aut}(V)$, the automorphisms of $V$.
A: What a homomorphism is always depends on the structure you're dealing with. The tricky part is that objects we consider to be a specific thing can carry multiple different kinds of structure, and what a homomorphism is depends on this structure. Let's take the complex numbers, for instance:
$(\mathbb C,+)$ is a group. The map $(\mathbb C,+)\to(\mathbb R,+),~x+\mathrm iy\mapsto x$ is a group homomorphism.
$(\mathbb C,\cdot)$ is a monoid. The map $(\mathbb C,\cdot)\to(\mathbb R,\cdot),~x+\mathrm iy\mapsto x^2+y^2$ is a monoid homomorphism. But it's not a group homomorphism between the additive groups from before.
$(\mathbb C,+,\cdot)$ is a ring. The map $(\mathbb C,+,\cdot)\to(\mathbb R^{2\times 2},+,\cdot),~x+\mathrm iy\mapsto\begin{pmatrix}x&-y\\y&x\end{pmatrix}$ is a ring homomorphism.
$(\mathbb C,\tau)$ with the usual topology $\tau$ is a topological space. The map $(\mathbb C,\tau)\to(\mathbb C,\tau),~z\mapsto\vert z\vert+\mathrm i z$ is a continuous map (= topological space homomorphism).
$\mathbb C$ (with some cumbersome to describe structure) is a real vector space. The map $\mathbb C\to\mathbb C,~z\mapsto \overline z$ is an $\mathbb R$-linear map (= real vector space homomorphism).
It's also a complex vector space. But the linear map from before is not a $\mathbb C$-linear map (= complex vector space homomorphism). However, $z\mapsto \mathrm iz$ is.
Again with some tediously described structure, $\mathbb C$ is a conformal manifold. The map $\mathbb C\to\mathbb C,~z\mapsto\exp z$ is a (locally) conformal map (= conformal manifold homomorphism).
So you see, you must specify the structure you mean when talking about homomorphisms, because certain objects can carry a multitude of structures. This is especially true for compound structures like inner product spaces, or even worse, Hilbert spaces, which carry multiple smaller structures by definition (Hilbert spaces are vector spaces, inner product spaces, normed spaces, and Banach spaces all at the same time). But as long as you specify what kind of homomorphism you're taking about at some point, you're good.
