Let $E$ be the elliptic curve over $\mathbb{Q}$ defined by 
$$
y^2=x^3-1.
$$
Let $p$ be an odd prime congruent $-1$ modulo $3$. Then, 
this curve becomes supersingular after ''reduction'' modulo $p$, also denoted by $E$. Note that the group of automorphisms of $E$, $\text{Aut} (E)$, is cyclic of order $6$. Let $\rho$ be a generator of $\text{Aut} (E)$.


Let $q=3k+1$ be another odd prime (for some $k\geq 2$). Then, $E$ has $(q+1)=(3k+2)$ distinct cyclic subgroups of order $q$. Let $\{C\_1, \dots, C\_{3k+2}\}$ be the complete list of these cyclic subgroups. Then, there are exactly two $i$ such that 
$$
\rho C\_i = C\_i.
$$
Let $D\_1$, $D\_2$ be two such subgroups. Then, for $C\_i \not\in \{D\_1, D\_2\}$, $\rho C\_i \neq C\_i$. Thus, $(E, C\_i), (E, \rho(C\_{i}))$ and $(E, \rho^2(C\_{i}))$ are isomorphic pairs. (Here, two pairs $(E, C)$ and $(E, D)$ are isomorphic if there is an automorphism sending $C$ to $D$.) By reordering them, we may assume that
$$
\rho C\_i = C\_{i+k} \text{ for } 1\leq i \leq 2k, \quad C\_{3k+1}=D\_1 \quad\text{ and } \quad C\_{3k+2}=D\_2.
$$
Then, the complete list of the isomorphism classes of pairs $(E, C)$ (with same $E$) is
$$
S:=\{[(E, D\_1)], [(E, D\_2)], [(E, C\_i)] : 1 \leq i \leq k\}.
$$
For $1\leq i \leq k$, $\text{Aut}([(E, C\_i)]) \simeq \{ \pm 1 \}$ and $\text{Aut}([(E, D\_1)]) \simeq \text{Aut}([(E, D\_2)]) \simeq \langle \rho \rangle$.


Any element of $S$ can be regarded as a point of $X\_0(q)(\overline{\mathbb{F}}\_p)$, where $\overline{\mathbb{F}}\_p$ is an algebraic closure of $\mathbb{F}\_p$, the finite field of $p$ elements. Some of them can be defined over $\mathbb{F}\_p$ and the others are not. 


Now I am asking the following: As a point of $X\_0(q)(\overline{\mathbb{F}}\_p)$, is $[(E, D\_1)]$ defined over $\mathbb{F}\_p$? i.e., $[(E, D\_1)] \in X\_0(q)(\mathbb{F}\_p)$? (namely, $\text{Frob}\_p(D\_1)=D\_1$? Here $\text{Frob}\_p$ is the Frobenius endomorphism in characteristic $p$.)
Similarly, is $[(E, D\_2)]$ defined over $\mathbb{F}\_p$?



Supersingular elliptic curves with automorphisms of order 6
No. Frobenius switches those two subgroups. 


To prove this, consider the eigenvalue of $\rho$ on each of those subgroups, which is one of the two sixth roots of unity in $\mathbb F\_q$. Because $\operatorname{Frob}\_p(\zeta)=\zeta^{-1}$ for $\zeta$ a third root of unity, we have $\operatorname{Frob}\_p (\rho)=\rho^{-1}$, and so Frobenius does not preserve this invariant and hence does not preserve the subgroup.


