What is the difference between a isomorphism of representations and a $G$-module homomorphism? Let $G$ be a finite group and $V,W$ be vector spaces. Let $\rho: G \to GL(V)$ and $\tilde{\rho}: G \to GL(W)$ be two representations of $G$. According to this answer, $\rho$ and $\tilde{\rho}$ are said to be isomorphic if there exists an isomorphism $A: V \to W$ such that $A(\rho(g)v) = \tilde{\rho}(g)(Av)$. On the other hand, I thought this map $A$ was called (at least in this context) a $G$-module homomorphism. Am I wrong or these are just different names for the same object?
 A: It all boils down to how much importance one gives to the  action of some algebraic structure on another.
The scalar multiplication in a vector space is perhaps the  example we least pay attention to,  so we just write
"$\lambda v$",
a notation that does not    even acknowledge the  operation being applied.
One could,  of course,   view  scalar multiplication  as an
action of the scalar field $K$ on the (so far only seen as an abelian) group $V$,  in which case we would consider a map
$$
  \Phi :K\to \text{End}(V),
  $$
effecting the scalar multiplication.
The old $\lambda v$ would then be given the pompous notation $\Phi (\lambda )(v)$.  Associativity of multiplication would in turn be
written as
$$
  \Phi (\lambda _1\lambda _2)(v) =   \Phi (\lambda _1)\big ( \Phi (\lambda _2)(v)\big ).
  $$
When a student  first encounters the concept of a group representation,  it is wise to denote it prominently.  However,  as one gets
used to it,  a lighter notation makes formulas a lot less cumbersome!
If we see a representation $\rho :G\to GL(V)$ as giving $V$ the structure of a $G$-module, then we are prone
to writing the action of $g$ on a vector $v$ a "$gv$",  rather than "$\rho (g)(v)$".
If $\tilde \rho :G\to GL(W)$  is another representation, then
a bijective linear map
$$
  A:V\to W,
  $$
satisfies  the condition
referred to by the OP, namely,
$$
  A(\rho(g)v) = \tilde{\rho}(g)(Av)
  \tag 1
  $$
iff
$$
  A(gv) = gA(v),
  \tag 2
  $$
which is part of the  definition  of a module map.
In conclusion, a bijective linear map   $A:V\to W$ establishes an isomrphism between the representations $\rho $ and $\tilde
\rho $ (i.e.,   satisfies (1)), if and only if it is a $G$-module map (i.e.,  satisfies (2)).
A: Given a group $G$ we can form two new structures:

*

*for a given field $k$, a $k$-representation of $G$, which is a pair $(V,\rho)$ where $V$ is a $k$-vector space and $\rho$ is a group homomorphism from $G$ to the set of all linear isomorphisms from $V$ to $V$;

*a $G$-module, which is a pair $(M,\mu)$ where $M$ is an (additive) abelian group and $\mu$ is a group homomorphism from $G$ to the set of all group isomorphisms from $M$ to $M$.

Remark: Forgetting the scalar multiplication, every representation of $G$ is naturally a $G$-module.
Now, if $(V_1,\rho_1)$ and $(V_2,\rho_2)$ are two $k$-representations of $G$, we can define a $k$-representation homomorphism $\alpha : (V_1,\rho_1) \to (V_2,\rho_2)$ as a linear map $\alpha : V_1 \to V_2$ such that $\alpha \circ \rho_1(g) = \rho_2(g) \circ \alpha$ for all $g \in G$. Note that if $\alpha$ is a linear isomorphism, this coincides with your notion of isomorphic representations!
Similarly, if $(M_1,\mu_1)$ and $(M_2,\mu_2)$ are two $G$-modules, we can define a $G$-module homomorphism $\beta : (M_1,\mu_1) \to (M_2,\mu_2)$ as a group homomorphism $\beta : M_1 \to M_2$ such that $\beta \circ \mu_1(g) = \mu_2(g) \circ \beta$ for all $g \in G$.
In conclusion, every representation homomorphism is also a $G$-module homomorphism in an obvious way, but not in the other way around. Hope this clarifies a bit.
Note: Similar constructions apply when we move from vector spaces or abelian groups to just sets, modules over a given ring, and topological spaces!
