 Solution

The Artin fixed-field theorem says that for a finite group $G$ of distinct automorphisms of a field $K$, the extension $K/K^G$ is Galois extension and has degree $|G|$. Here $y=x^p-x$ is invariant under translations because $(x+a{)}^p-(x+a)=x^p-x$ for $a\in {\mathbb{F}}_p$. Hence ${\mathbb{F}}_p(y)\subset K^G$. Also $x$ is a root of $T^p-T-y$, so $[K:{\mathbb{F}}_p(y)]\le p$. The fixed-field theorem gives $[K:K^G]=p$, forcing $\boxed{K^G={\mathbb{F}}_p(y)}$. The extension is separable; explicitly the derivative of $T^p-T-y$ is $-1$.

For the affine substitutions, $y$ transforms to $dy$, since $d^p=d$ and $a^p=a$. Therefore $z=y^{p-1}$ is fixed by all $p(p-1)$ substitutions. The polynomial $(T^p-T{)}^{p-1}-z$ has $x$ as a root and degree $p(p-1)$. Apply the same degree comparison and the Artin fixed-field theorem to obtainMoreover $[{\mathbb{F}}_p(y):{\mathbb{F}}_p(z)]=p-1$, so the minimal polynomial of $y$ is $\boxed{T^{p-1}-z}$. It splits as ${\prod }_{d\in {\mathbb{F}}_p^{\times }}(T-dy)$, and its derivative is nonzero at every root. Thus $K^G/K^H$ is a Galois extension with group ${\mathbb{F}}_p^{\times }$, acting by $y\mapsto dy$.

There is a harmless convention issue in labeling the substitutions: $f(x)\mapsto f(dx+a)$ composes in the reverse order to the displayed affine matrices. It is naturally a right action on functions, or a left action after inserting inverses. The set of automorphisms and all the fixed fields just computed are unchanged by that convention.