What is the $\pi_1$-action on the hom-sheaf between two finite etale covers? Say you have two finite etale covers $X\rightarrow S$, $Y\rightarrow S$. The hom sheaf $\mathcal{H}om_S(X,Y)$ on the etale site $\text{Sch}/S$ is finite locally constant, hence representable by some scheme $H$, also finite etale over $S$. Let $s\in S$ be an arbitrary geometric point, then the fiber $H_s$ is essentially just the set of all functions from $X_s\rightarrow Y_s$ (as sets).
My question is, what is the action of $\pi_1(S,s)$ on $H_s$?
My guess is this: for $\gamma\in\pi_1(S,s)$, and $h\in H_s$ a function from $X_s\rightarrow Y_s$, $\gamma.h := \gamma\circ h\circ\gamma^{-1}$, where $\gamma$ is viewed as an automorphism of the set $Y_s$, and $\gamma^{-1}$ is viewed as an automorphism of the set $X_s$.
Is this correct?
Apparently this is what I computed a long time ago, but now I can't for the life of me remember how I computed this or why I thought it was correct.
 A: Yes, this is correct. Notice that your hom sheaf is the internal hom in the category of étale sheaves on $S$ resp. the subcategory of finite étale schemes over $S$. Recall that the category of finite $\pi_1(S,s)$-sets is equivalent to the category of finite étale schemes over $S$. Using the equivalence, the problem becomes simply: If $X,Y$ are finite $G$-sets for a (profinite) group $G$, how does $G$ act on the set of maps $\mathrm{Map}(X,Y)$ in such a way that this becomes the internal hom in the category of finite $G$-sets? Well there is only one way of doing this, as you said, we define the action of $g \in G$ on $h : X \to Y$ by $(gh)(x)=g \cdot h(g^{-1} \cdot x)$. This is the internal hom, i.e. for every finite $G$-sets $Z$ there is a natural bijection $\hom_G(Z,\mathrm{Map}(X,Y)) \cong \hom_G(Z \times X,Y)$. In fact, currying gives a natural bijection $\mathrm{Map}(Z,\mathrm{Map}(X,Y)) \cong \mathrm{Map}(Z \times X,Y)$, and a map $f : Z \times X \to Y$ is $G$-equivariant iff $f(gz,gx)=g f(z,x)$ (for all $g,z,x$) iff $f(gz,x)=g f(z,g^{-1} x)$ (for all $g,z,x$) iff $z \mapsto f(z,-)$ is $G$-equivariant.
