Let $\overline{\mathbb{F}_p}=\bigcup_{n\geqslant 1} \mathbb{F}_p$. We know that $\mathbb{F}_{p^m}\subset \mathbb{F}_{p^n}$ iff $m\mid n$, and $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)=\langle (F:x\mapsto x^p)\rangle$.
Now let $\sigma$ be an automorphism of $\overline{\mathbb{F}_p}$, then we can restrict
$$\sigma|_{\mathbb{F}_{p^m}}=F^{a_m}$$
and
$$\sigma|_{\mathbb{F}_{p^n}}=F^{a_n}.$$
My book now claims without proof that there is a certain divisbility criterium to let this be compatible, that is: $$a_n\equiv a_m\pmod{m} \quad\text{ if }m\mid n.$$
I am trying to prove this claim by writing it out in order to understand this. I know that $F^{a_n}:x\mapsto x^{p^{a_n}}$ and $F^{a_m}:x\mapsto x^{p^{a_m}}$. Now $F^{a_n}\mid _{\mathbb{F}_{p^m}}=F^{a_m}$, so for all $x\in\mathbb{F}_{p^m}$, we must have $x^{p^{a_n}}=x^{p^{a_m}}$. In $\mathbb{F}_{p^m}$, $x^{p^m}=(x^{p^m})^{p^m}=\ldots=x$, so $x^{p^{a_n}}=x^{kp^{m}\cdot p^{a_m}}$ ...
Is this correct? How should I prove this?
$F$ is an automorphism of $\mathbb{F}_{p^m}$.
So if $F^{a_n} = F^{a_m}$, we have $a_n = a_m$ modulo the order of $F$ in $Aut(\mathbb{F}_{p^m})$. But the order of $F$ is $m$.
The thing that works here and that you didn't see is that you can actually divide by $p$ in the exponents, because you can compose by $F^{-1}$.