This question is related to an exercise in Bosch's Algebraic Geometry (Chap. 4.5, Ex. 2)
Fix a group $G$ and write $\textbf{G}$ for the one-point category with morphisms given by the group $G$ and $\textbf{Ab}$ for the category of abelian groups. The claim is that ${\rm Hom}(\textbf{G},\textbf{Ab})$ is the category of $G$ - modules, i.e. $\mathbb{Z}$-modules with a $G$ - action.
For clarification, given categories $C_1,C_2$, ${\rm Hom}(C_1,C_2)$ is the category of all covariant functors $C_1\rightarrow C_2$, where a morphism $\phi:F\rightarrow G$ between two such functors is a functorial morphism.
These are my thoughts so far: A functor $F:\textbf{G}\rightarrow \textbf{Ab}$ defines an abelian group $F(G)$. I want to define and action $g\cdot x$ for $g\in G$ and $x\in F(G)$ induced by $F$, but I don't know how. What I can define is an action of ${\rm Hom}_G(G,G)$ on $F(G)$ by $f\cdot x:= F(f)(x)$ for a morphism $f:G\rightarrow G$ and $x\in F(G)$. But this is not what I want.
Can anyone give me a hint?
Don't call $G$ both the group and the unique object of $\bf G$: a nameless label like $*$ is better, especially because it is absolutely irrelevant "what" is $*$.
Call $X=F*$.
A morphism $g \in {\bf G}(*,*)$ is an element of $G$. By the definition of $\bf G$, $g$ is an endomorphism of $*$. Actually, an auto morphism.
Every such $g$ induces an endofunction $F(g) : F* \to F*$, that has to be invertible because $g$ was invertible ($g^{-1}$ is its inverse). This defines a function $a : G \to {\rm Aut}(X)$. The fact that $F$ was a functor implies that this function is a group homomorphism, so $G$ acts on $X$. Actually, $F$ is a functor if and only if $a$ is an action.