Pullback of a vector bundle extension class

Question

Let $ X $ be a variety with an automorphism $ \phi : X \rightarrow X $. Suppose there is a short exact sequence of vector bundles $ 0 \rightarrow F \rightarrow E \rightarrow G \rightarrow 0 $ on $ X $ such that there are isomorphisms $ f: F \rightarrow \phi^* F $ and $ g : G \rightarrow \phi^* G $. Is there any reason for there to exist an isomorphism $ e : E \rightarrow \phi^*E $ such that the obvious diagram commutes

$$\require{AMScd} \begin{CD} 0 @>{}>> F @>{p}>> E @>{q}>> G @>{}>> 0\\ @. @V{f}VV @V{e}VV @V{g}VV @.\\ 0 @>{}>> \phi^*F @>{\phi^*p}>> \phi^*E @>{\phi^*q}>> \phi^*G @>{}>> 0 \end{CD}$$

If not, what is a counterexample, preferably on $ X $ projective?

Answer 1

For a counterexample, one can take $X$ an elliptic curve over a field of characteristic not $2$, $\phi$ the involution sending each point to its inverse under the group law, $F$ and $G$ both $\mathcal O_X$, $f$ and $g$ the obvious isomorphisms, $E$ any nontrivial extension.

If $E$ corresponds to a class $\alpha \in H^1(X, \mathcal O_X)$, then $\phi(E)$ corresponds to $\phi(\alpha)=-\alpha \neq \alpha$, since $\alpha\neq 0$, so $\phi(E)$ is not isomorphic to $E$ by an isomorphism making that diagram commute.