Why are equivariant morphisms of $G$-torsors necessarily isomorphisms?

Question

This was something I read on the Stacks project, but whose proof was omitted.

Simply stated, if $f\colon E\to F$ is a $G$-equivariant morphism of $G$-torsors over a scheme $X$, why is $f$ necessarily an isomorphism?

Answer 1

By definition, after base changing to a suitable cover of $Y$ of $X$ depending on the topology, the $G$-torsors become trivial, so that $E_Y$ and $F_Y$ both become isomorphic to $Y \times G$. Now any equivariant map $f': Y\times G \rightarrow Y\times G$ over $Y$ is going to be multiplication by some element of $G$ on the second factor, hence an isomorphism. This isomorphism then descends back down to $f$.