Group action on group action morphisms

Question

Let’s assume $G$ is a group. Let $BG$ be the category with one object which its morphisms form a group isomorphic to $G$. Now, note that for me if $\mathcal{C}$ is an arbitary category, a $G$-action is a functor $F:BG\to \mathcal{C}$ (When, $\mathcal{C}=Set$, we will have a usual $G$-action). Now, first of all my question is that can we put a $G$-action on $Hom_{\mathcal{C}^{BG}}(F_1,F_2)$? i.e. can we put a $G$ action on the morphisms of $G$-actions? My feeling is that the answer is yes and my main question is that if the answer is yes then is it doable in categorical terms and is it possible to generalize it? For example, when $R$ is a commutative ring, we can look at $R$-modules as functors from $BR$ to the category of abelian groups and if M, N are two are $R$-modules, then we can look at $Hom(M,N)$ also as an $R$-module.

Answer 1

I believe that @Jackozee Hakkiuz wanted you to realize that there is an action only if $R$ is commutative, and the same hold for group actions : you cannot expect a natural $G$-action on the group of morphisms of $G$-sets unless $G$ is abelian.

Note however, that you can put a $Z(G)$-action and this can be constructed categorically : let $\operatorname{id}_{BG}$ then identity functor of the category $BG$ and note that the group of natural transformations $\operatorname{id}_{BG}\to\operatorname{id}_{BG}$ is the center of $G$, in other words : $Z(G)=\operatorname{Nat}(\operatorname{id}_{BG},\operatorname{id}_{BG})$. Then, if $F_1:BG\to \mathcal{C}, F_2:BG\to \mathcal{C}$ are to $G$-object in $\mathcal{C}$, then write $F_1=F_1\circ\operatorname{id}_{BG}$ and $F_2=F_2\circ\operatorname{id}_{BG}$ and you have a natural action of $Z(G)$ by "horizontal" composition.