The affine group AGL(1,p) has one kernel-conjugacy class of complements to its translation subgroup

If $p=2$, then ${\mathbb{F}}_2^{\times }=\{1\}$ is trivial, so the only crossed homomorphism ${\mathbb{F}}_2^{\times }\to {\mathbb{F}}_2$ is the zero map. Hence $H^1({\mathbb{F}}_2^{\times },{\mathbb{F}}_2)=0$ by [L4].

Suppose $p>2$. By [L2] and [L3], choose a generator $u$ of the cyclic group ${\mathbb{F}}_p^{\times }$. Any crossed homomorphism $z:{\mathbb{F}}_p^{\times }\to {\mathbb{F}}_p$ is determined by $m=z(u)$, so it suffices to show that $m$ is always principal.

Because $p>2$, the generator $u$ is not $1$, so $u-1\mathrm{\ne }0$ in the field ${\mathbb{F}}_p$. By [L2] it is therefore invertible. Choose $a\in {\mathbb{F}}_p$ with $(u-1)a=m$. Then the principal cocycle $g\mapsto g\cdot a-a$ agrees with $z$ on the generator $u$, hence everywhere by [L3]. Thus every crossed homomorphism is principal, and [L4] gives $H^1({\mathbb{F}}_p^{\times },{\mathbb{F}}_p)=0$.

Steps 1.1 and 2.1 show that $H^1({\mathbb{F}}_p^{\times },{\mathbb{F}}_p)=0$ for every prime $p$. Now [L1] shows that there is exactly one ${\mathbb{F}}_p$-conjugacy class of complements to the translation subgroup in $\mathrm{AGL}(1,p)$.