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?



Pullback of a vector bundle extension class
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.


