Let $X$ be an algebraic variety (i.e. an integral separated scheme of finite type over an algebraically closed field $k$) and let $G$ be a finite group of automorphisms of $X$. Suppose (as we may in the case of quasi-projective varieties) that for any $x$ in $X$, the orbit $G_x$ of $x$ is contained in an affine open subset of $X$. A classical result states that there exists a (unique) $k$-variety $Y$ together with a finite, surjective and separable morphism $\pi \colon X\to Y$ such that:
1) As topological space $(Y,\pi)$ is the quotient of $X$ for the action of $G$.
2)There is a natural isomorphism $\mathcal{O}_Y \to \pi_{*}(\mathcal{O}_X)^{G}$.
In this setting, let $\mathcal{F}$ be a coherent sheaf on $Y$. Since for any $g$ in $G$ we have a commutative diagram:
\begin{array}{ccc} X & \to^{g} & X \\ \downarrow^{\pi} & & \downarrow^{\pi}\\ Y & \to^{id} & Y \end{array}
there should be a map between $\pi^{*}\mathcal{F}$ and $\pi^{*}\mathcal{F}$ induced by $g$, that is a natural automorphism. However, I can't really understand what the map is supposed to do... The conclusion is that $G$ acts on the pullback of the sheaf $\mathcal{F}$ in a compatible manner (with respect to the action on $X$), but I can hardly imagine what is really happening here, beyond the formal arguments. Can anybody explain this in a more concrete way?
P.S. Sorry for the bad $\TeX$ typesetting: I suppose that the above diagram should be understood as a commutative triangle.