Let $p$ be a prime greater than 3, and let $M$ be a positive integer prime to $p$. Let $\Sigma$ be the set of isomorphism classes of pairs $(E, C\_M)$, where $E$ is a supersingular elliptic curve over $\overline{\mathbb{F}\_p}$ and $C\_M$ a cyclic subgroup of $E$ of order $M$. Then, by Deligne-Rapoport, $X\_0(pM)\_{\overline{\mathbb{F}\_p}}$ has two components isomorphic to $X\_0(M)\_{\overline{\mathbb{F}\_p}}$, and they meet transversally at "the points'' in $\Sigma$. 


Let $\Sigma\_4$ (resp. $\Sigma\_6$) be the subset of $\Sigma$ which consists of pairs $(E, C\_M)$ such that $\# \text{Aut} (E, C\_M)=4$ (resp. $6$). If $M=1$ then an element in $\Sigma\_4$ or in $\Sigma\_6$, if it exists, is defined over $\mathbb{F}\_p$. 


My question is as follows: Suppose that $M>1$. Let $s=(E, C\_M) \in \Sigma\_6$ (or in $\Sigma\_4$). I guess $s$ is not defined over $\mathbb{F}\_p$ (but only in $\mathbb{F}\_{p^2}$), but I cannot prove this. I hope someone in MO knows the answer.



Supersingular elliptic curves with extra automorphisms
If there is a supersingular elliptic curve $E / {\bf F}\_p$ with
an automorphism of order $4$ or $6$, then the automorphism is not
defined over ${\bf F}\_p$, but plenty of cyclic subgroups *are*
defined over ${\bf F}\_p$. This is because the Frobenius endomorphism $\varphi$
generates an imaginary quadratic ring and there are many choices of
$a, b \in \bf Z$ with $a \not\equiv 0 \bmod p$ for which the endomorphism
$\alpha = a + b\varphi$ has cyclic kernel. (The condition on $a$ assures that 
$\alpha$ is separable.)


