Finite Field Extensions Are Cyclic Galois

Equivalently, $\mathrm{Fr}_p^n=\mathrm{id}$ on $\mathbb{F}_q$ and the smallest positive integer $m$ with $\mathrm{Fr}_p^m=\mathrm{id}$ is $m=n$. In particular, by degree = group order one has $[\mathbb{F}_q:\mathbb{F}_p]=n$.

More generally, if $\mathbb{F}_{p^m}\subseteq \mathbb{F}_{p^n}$ (so $m\mid n$), then $\mathbb{F}_{p^n}/\mathbb{F}_{p^m}$ is Galois with cyclic group generated by $x\mapsto x^{p^m}$, and the fixed field of $\langle \mathrm{Fr}_p^m\rangle$ is exactly $\mathbb{F}_{p^m}$ (compare the Galois correspondence).

Examples

1. $\mathbb{F}_4/\mathbb{F}_2$ (order 2 group). Realize $\mathbb{F}_4=\mathbb{F}_2(\alpha)$ with $\alpha^2+\alpha+1=0$, i.e. $\mathbb{F}_4\cong \mathbb{F}_2[t]/(t^2+t+1)$. Then Frobenius is $x\mapsto x^2$. It fixes $\mathbb{F}_2$ and sends
   $$\alpha \longmapsto \alpha^2 = \alpha+1,$$
   which is nontrivial; hence $\mathrm{Gal}(\mathbb{F}_4/\mathbb{F}_2)=\{ \mathrm{id},\, \mathrm{Fr}_2\}$ is cyclic of order $2$.

1. $\mathbb{F}_9/\mathbb{F}_3$ (order 2 group, explicit action). Take $\mathbb{F}_9=\mathbb{F}_3(\beta)$ with $\beta^2+1=0$ (irreducible over $\mathbb{F}_3$), so $\beta^2=-1=2$. Frobenius is $x\mapsto x^3$, and
   $$\beta \longmapsto \beta^3=\beta\cdot \beta^2 = 2\beta.$$
   Applying Frobenius twice sends $\beta\mapsto (2\beta)^3=8\beta^3\equiv 2\cdot(2\beta)=4\beta\equiv \beta$ mod $3$, so the Frobenius automorphism has order $2$, as predicted.

1. Subfields in $\mathbb{F}_{p^6}$. The group $\mathrm{Gal}(\mathbb{F}_{p^6}/\mathbb{F}_p)$ is cyclic of order $6$ generated by $\mathrm{Fr}_p$. The subgroup $\langle \mathrm{Fr}_p^2\rangle$ has order $3$, and its fixed field is the unique intermediate field of size $p^2$, namely $\mathbb{F}_{p^2}\subset \mathbb{F}_{p^6}$.