Automorphism group of the general affine group of the affine line over a finite field? I am wondering what the structure of the automorphism group of the general affine group of the affine line over a finite field looks like. I'll make that a bit more precise:
If $k$ is a finite field, and $\operatorname{AGL}_1(k)$ its group of affine transformations, i.e. maps of the form
$$k\ \longrightarrow\ k:\ x\ \longmapsto\ ax+b,$$
with $a\in k^{\times}$ and $b\in k$, then what is the isomorphism type of $\operatorname{Aut}(\operatorname{AGL}_1(k))$?
I know that $\operatorname{AGL}_1(k)\cong k\rtimes k^{\times}$, where the semi-direct product is given by the natural action of $k^{\times}$ on $k$ by multiplication. Also, as the center of $\operatorname{AGL}_1(k)$ is trivial, it is isomorphic to a subgroup of its isomorphism group. Any automorphism of $\operatorname{AGL}_1(k)$ restricts to a group automorphism of $k^{+}$, of which there are very many, unfortunately.
What is a good way to approach this problem?
 A: Let $G = {\rm AGL}_1(k)$. Then ${\rm Aut}(G) = {\rm A \Gamma L}_1(k) \cong G\langle \gamma \rangle$, where $\gamma$ is a generator of the group of field automorphisms of $k$. So if $|k|=p^e$ with $p$ prime, then $|\gamma|=e$ and you $\gamma:x \mapsto x^p$ for $x \in k$.
Here is a sketch proof. Any $\alpha \in {\rm Aut}(G)$ must fix the normal subgroup $k$ of the semidirect product and, since the complements are all conjugate in $G$, we can assume (by multiplying $\alpha$ by an inner automorphism) that if fixes the principal complement $k^\times$. Since $k^\times$ acts transitively by conjugation on $k \setminus \{0\}$, by multiplying $\alpha$ by an inner automorphism again, we can assume that $\alpha(1)=1$.
For $0 \ne a \in k  < G$, I will use $\bar{a}$ to denote the corresponding element of the complement $k^\times$. So the semidirect product action is $\bar{a}b\bar{a}^{-1} = ab$.
Now, for $0 \ne a \in  K$, using $\alpha(1)=1$, we have
$$\alpha(a) = \alpha(\bar{a}1\bar{a}^{-1}) =\alpha(\bar{a})1\alpha(\bar{a})^{-1},$$
so $\overline{\alpha(a)} = \alpha(\bar{a})$. In other words $\alpha$ is acting in the same ways on $k -\{0\}$ and on $k^\times$.
So, for $a,b \in k \setminus \{0\}$,
$$\alpha(a)\alpha(b) = \overline{\alpha(a)} \alpha(b) \overline{\alpha(a)} ^{-1} = \alpha(\bar{a}) \alpha(b)\alpha(\bar{a})^{-1} = \alpha(\bar{a}b\bar{a}^{-1}) = \alpha(ab)$$
and hence $\alpha$ is acting as a field automorphism of $k$.
