Let $K/\mathbb F\_q(x)$ be a finite Galois extension with Galois group $G$. Let $Aut(K)$ be the group of $\mathbb F\_q$-automorphisms of $K$.
Obviously, $G\subseteq Aut(K)$. It is well known that
$H^1(G,K^\*) = 1$ [Hilbert 90]. But does the following hold: $H^1(Aut(K), K^\*)=1$?


Thanks in advance.



Variant of Hilbert 90 for Galois extensions
This is an application of Artin's Lemma to reduce to Hilbert 90. Namely, let $K$ be any field at all, and $G$ any *finite* subgroup of ${\rm{Aut}}(K)$ (as noted in this comments, such finiteness holds for the $K$ in the question with $G = {\rm{Aut}}(K)$). Then by Artin's Lemma the field $K$ is finite Galois over its subfield $K^G$ of $G$-invariants with $G \rightarrow {\rm{Gal}}(K/K^G)$ an isomorphism. Hence, ${\rm{H}}^1(G,K)$ vanishes by Hilbert 90.


(In the geometric setting of the question, $K^G$ is the function field of the quotient $C/G$ where $C$ is connected regular proper $\mathbf{F}\_p$-scheme of dimension 1 with function field $K$.)


