In what follows, let $G$ be a finite group, $F$ a field and $V$ a vector space over $F$. First we fix some definitions:
Representation: A homomorphism $\phi:G\to GL(V)$.
In this sense, if $V$ is a finite-$n$-dimensional vector space over $F$, we can conflate $\rho$ with the image of $\rho(G)$ a finite collection of $n\times n$ matrices over $F$ corresponding to the action of $g$ on $v\in V$. Sometimes, we will conflate $\rho(G)$ with $V$ itself.
Subrepresentation: If $W$ is a subspace of $V$ then we say $W$ is a subrepresentation of $V$ if it is $G$-invariant. i.e. the action of $g\in G$ on vectors in $W$ map to other vectors in $W$.
It is clear to me from this that $V,\{0\}$ are $G$-invariant.
Now, my confusion with these concepts arises when trying to justify why we can conflate $V$ with $\rho(G)$. This becomes particularly apparent when I try to prove Schur's lemma.
First, I assume the following form of Maschke's Theorem:
If $V$ is a representation of some finite group (by which we mean that there is a $rho:G\to GL(V)$), and there exists $U$ a subrepresentation of $V$ (taken to mean that there is a $G$-invariant subspace $U$ of $V$), then there is another subrepresentation $W$ of $V$ so $V=U\oplus W$. i.e. $V$ is the direct sum of $U, W$.
The version of Schur's lemma I am trying to understand in light of the above is as follows:
If $M,N$ are irreducible representations of $G$ and $\phi$ is a $G$-linear transformation in the sense that $\phi$ commutes with the action of $g\in G$ on vectors $V$ in the representation $M$, then $\phi$ is either the zero map or an isomorphism. Furthermore, $\phi$ must take the form $\lambda I_V$ for $\lambda\in F$.
Here is proof attempt/understanding of the proof:
We will show that $\ker{\phi},\text{Im}{\phi}$ are $G$-invariant since they are subspaces of $V, W$ respectively, where $V,W$ are the vector spaces we conflate with the representations $M,N$. Let $v\in\ker{\phi}$. Then $\phi(v)=0$. So $\phi(gv)=\rho(g)\phi(v)=\rho(0)=0$, as required. Similarly for the image, let $y\in\text{Im}{\phi}$ then there is $x\in V$ so $\phi(x)=y$. So $\phi(gx)=g\phi(x)=\rho(g)y$. I think in this context $\rho$ is associated with $N$ since otherwise $\rho(g)y$ makes no sense. Please correct me if I am wrong. But anyway, this immediately implies that the kernel and image are subrepresentations. By the irreducibility of $M,N$ we conclude that $\phi$ is an isomorphism.
A question which occurs to me immediately at this point in the proof is the necessity of all of this. i.e. could we not also prove this statement by the following reasoning: if $\phi$ is linear from $V\to W$, then $V,W$ have the same dimension, otherwise the kernel fails to be non-trivial? Why the need to show $G$-invariance? Does this statement alone not already give us the fact that $\phi$ is bijective and hence an isomorphism since certainly it is a homomorphism by its linearity which is assumed?
As for the second bit of the proof, the logic seems to go completely over my head.
We would like that show that $\phi$ takes the form $\lambda I_V$. Now, in particular, I guess that $\phi$ is really a homomorphism of linear transformations since $M,N$ are representations which are simply a collection of transformations on some ambient vector spaces $V,W$. i.e. It seems to me that $\phi$ maps elements of $L(V,V)$ onto elements of $L(W,W)$. Am I mistaken in thinking this? Now since we just established that $V$ and $W$ have the same cardinality, $L(V,V)$ should be isomorphic to $L(W,W)$? I guess in some sense since $\phi$ is a map from $L(V,V)\to L(W,W)$, we can identify it with a map from $V\to W$? Is this line of reasoning correct? If so, then I guess if $\phi$ was not of the form $\lambda I_v$ for some lambda, then it would have a non-trivial proper eigenspace in which case $W$ wouldn't be irreducible? Is this moving in the right direction?
I guess my largest concern lies in understanding the justification behind why we can conflate all of these objects. Apologies for the long-winded question. Any help clearing all this up for me is immensely appreciated as I've been muddling around with it for a good while now. Thanks in advance!
Edit: As a comment kindly pointed out, I should specify $F$ is an algebraically closed field. I figured out my point of confusion thanks to the poster below. As for the second part of Schur's Lemma, a proof is as follows. If $\phi$ is a $G$-linear map over a complete field, it has eigenvalues in this field. Say one of them is $\lambda$. Then consider the transformation $\phi-\lambda I$. If this does not have the trivial kernel, then since $V$ is irreducible the kernel must be all of $V$. i.e. $\phi-\lambda I=0$. The result follows.
Best Answer
I would say we're not conflating these objects, we're conflating which of them gets referred to as a "representation" as a shorthand. Recall a representation (of a group $G$) is formally a pair $(V,\rho)$ consisting of a vector space $V$ and a group homomorphism $G\to\mathrm{GL}(V)$. (In more advanced representation theory, you can talk about $G$ being topological or Lie and what that means for $\rho$ too.) For convenience, we often call $V$ or $\rho$ the "representation" with the other just assumed to be part of the mix.
By the way, do be careful about conflating $\rho$ and $\rho(G)$. The latter is the range of the homomorphism, not the homomorphism itself. Inequivalent representations can have the same range, plus if $\rho$ isn't faithful (injective) then $\rho(G)$ can be strictly smaller than $G$.
In any case, it doesn't sound to me that this is your issue at all. It sounds to me is your issue is in understanding that the word "isomorphism" can mean different things in different contexts, and in particular an isomorphism of vector spaces is not necessarily an isomorphism of representations.
A bijective linear map is a vector space isomorphism but may not even be a morphism of representations to begin with, let alone an isomorphism (AKA equivalence) of representations.
Don't mix up which category of mathematical object we're talking about; you wouldn't (e.g.) automatically assume a bijection (an "isomorphism" in the category of sets) between topological spaces must be continuous, let alone a homeomorphism (an "isomorphism" in the appropriate category of topological spaces).
In the category of representations (of a given group, over a given field), a morphism of representations $(V_1,\rho_1)$ and $(V_2,\rho_2)$ is a linear transformation $\phi:V_1\to V_2$ which is "$G$-equivariant" or "intertwines" the $G$-action. Formally, this means that $\phi$ satisfies $\phi(\rho_1(g)v)=\rho_2(g)\phi(v)$ for all $g\in G,v\in V$.
No, this is all wrong.
That is the idea, yes, and you have to show the eigenspace would be a subrepresentation.