Let $A$ be an algebra and $f:A\to A$ an automorphism of $A$. Further, we denote the category of finite-dimensional left $A$-modules with $A$-mod.
$f$ defines a functor $F:A\text{-mod}\to A\text{-mod}$, which sends an $A$-module $V$ with action $\rho_V$ to the $A$-module $V$ with action $\rho'_V:=\rho_V\circ (f\times id)$ and a morphism $f\in A\text{-mod}$ to itself.
If I define a map $g:V\to V,v\mapsto \rho_V(a,v)$, for fixed $a\in A$, the map $F(g)$ is by definition equal to $g$.
My question is: What is $F(g)(v)$? Is it equal to $\rho_V(a,v)$ or is it equal to $\rho'_V(a,v)$?
I am a bit confused because at the one hand $F(g)$ is a map $F(V)\to F(V)$ which would make me think $\rho'_V(a,v)$ is correct but at the other hand $F(g)=g$ so I would guess $\rho_V(a,v)$ is the right choice.
The short answer is that $F(g) = g$, so $F(g)(v) = g(v) = \rho_V(a, v)$.
For the longer answer, first, note that for $b \in A$, $v \in V$, we have $$g(\rho_V(b, v)) = \rho_V(a, \rho_V(b, v)) = \rho_V(ab, v)$$ but $$\rho_V(b, g(v)) = \rho_V(b, \rho_V(a, v)) = \rho_V(ba, v)$$ and so unless we assume that $a$ is central in $A$, your map $g$ is not in general a module homomorphism.
Now assume $a$ is indeed central.
$$F(g)(\rho_V'(b, v)) = g(\rho_V(f(b), v)) = \rho_V(a, \rho_V(f(b), v)) = \rho_V(a \cdot f(b), v)$$
$$\rho_V'(b, F(g)(v)) = \rho_V(f(b), F(g)(v)) = \rho_V(f(b), \rho_V(a, v)) = \rho_V(f(b) \cdot a, v)$$
This confirms that $F(g)$ is indeed a morphism of the module $(V, \rho_V')$ to itself. (This has to be the case, for $F$ to be a functor, but since we're uncertain about the precise definition of the functor, it's worth checking for reassurance).