Order of an orbit of Frobenius action on a algebraically closed field of characteristic p Consider the action of the Frobenius homomorphism $F^{2}:\,\overline{\mathbb{F}_{q}}\rightarrow\overline{\mathbb{F}_{q}},\,x\rightarrow x^{q^{2}}$
  over $\overline{\mathbb{F}_{q}}$
 . Let $s=\left\{ s_{1},\ldots,s_{k}\right\}$ 
  be an orbit of this action such that $s=\tilde{s}$
 , where $\tilde{s}:=\left\{ s_{1}^{-q},\ldots,s_{k}^{-q}\right\}$ 
 . What about the order of the orbit $s$
 ? I suspect that it has to be odd. Thanks in advance for any suggestions or helps.

Your hunch is correct.
Assume that $s$ is the orbit of the element $t\in\overline{\Bbb{F}_q}$. The assumption implies that $t^{-q}\in s$, so there exists an integer $m$ such that
$$
t^{-q}=(F^2)^m(t)=t^{q^{2m}}.
$$
Raising both sides of this to power $q^{2m}$ shows that
$$
\begin{aligned}
t^{q^{4m}}&=(t^{-q})^{q^{2m}}\\
&=(t^{q^{2m}})^{-q}\\
&=(t^{-q})^{-q}\\
&=t^{q^2}.
\end{aligned}
$$
Therefore $t^{q^{4m-2}}=t$. If we denote $\Phi=F^2$ this means that the iterate $\Phi^{2m-1}(t)=t$. Therefore the size of the orbit $s$ is a factor of $2m-1$, i.e. an odd integer.
