= 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 obtain
$$
\boxed{K^H=\mathbb F_p(z),\qquad z=(x^p-x)^{p-1}.}
$$
Moreover $[\mathbb F_p(y):\mathbb F_p(z)]=p-1$, 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.
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