Consider the quotient stack $[X/G]$. Suppose $f: S' \to S$ morphism and $(E \to X, E \to S)$ is an object in $[X/G]$. Now we base change along $f$, and get $(f^*E :=E \times_S S' \to E, E \times_S S' \to S')$. Assume that $G$ acts on $E \times_S S'$ by acting only on the 1st factor.
We focus in particular on the morphism $E \times_S S' \to E$ and want to show that it is $G$-equivariant. (Since $E \to X$ is $G$-equivariant, then the composition of the two morphisms will yield that $E \times_S S' \to X$ is $G$-equivariant).
Question: 'Since $G$ acts trivially on $S'$, then $E \times_S S' \to E$ is $G$-equivariant'. Why is this statement true?
-Then one can regard $\alpha: E \to E$ as the morphism $E \times_S S' \to E$ (since $G$ acts trivially on $S'$). Then why would $\alpha(e \cdot g) = \alpha(e) \cdot g$?
-'Counter-example' : $e_1 \cdot g = e_2, e_2 \cdot g = e_1$ and $\alpha(e_1) = e_1, \alpha(e_2) = e_1$. Then, $\alpha(e_1 \cdot g ) = e_1 \neq e_2 = \alpha(e_1) \cdot g$