Let $p$ be prime and $q = p^n$. Let $E$ be an elliptic curve over $\mathbb{F}\_q$, and let $E^{(p)}$ be the pullback of $E$ by the $p$-power Frobenius of $\mathbb{F}\_q$. If $E$ is isomorphic (over $\mathbb{F}\_q$) to its Galois conjugate $E^{(p)}$, then does it follow that $E$ is the base change of an elliptic curve over $\mathbb{F}\_p$? If so, what is the argument?


Note that similar statements are not true over infinite fields: for instance, $\mathbb{Q}$-curves are isomorphic to all their Galois conjugates, yet need not descend to actual elliptic curves over $\mathbb{Q}$.



Is an elliptic curve that is isomorphic to its Frobenius conjugate defined over $\mathbb{F}_p$?
Almost, but not quite. If $j(E)$ is the $j$-invariant of $E$ then, under your hypothesis, $j(E)=j(E^{(p)})=j(E)^p$, so $j(E) \in \mathbb{F}\_p$. Hence $E$ is either defined over $\mathbb{F}\_p$ or is a twist of such a curve. If you take the quadratic twist in $\mathbb{F}\_{p^2}$ of an elliptic curve defined over $\mathbb{F}\_p$ you get a counterexample. 


