Let's $K$ be a finite field with $q=p^r$ elements where $p$ prime is the characteristic of $K$.
We denote $\rm{AGL}(n,q)$ the affine general linear group of dimension $n$ over $K$. I'm trying to identify $\rm{Aut}(\rm{AGL}(1,q))$.
$\rm{AGL}(n,q)$ is isomorphic to $K^* \times K$ equipped with the operation $$(a_1,b_1)(a_2,b_2)=(a_1 a_2,a_2 b_1 + b_2).$$ I computed the $m$-th power of an element $(a,b)$ which is $$(a,b)^m=\begin{cases} (1,mb) & \text{for } a=1\\ (a^m,b \frac{a^m-1}{a-1}) & \text{for } a \neq 1 \end{cases}$$ From there I induce that an element $(1,b)$ with $b \neq 0$ is having order $p$. The order of an element $(a,b)$ with $a \neq 1$ is equal to the order of $a$ in $K^*$.
But I'm having hard time to move from there...
Note: this is a question I raise following this one.