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 $K$.)

allcases. $\endgroup$ – Laurent Moret-Bailly May 27 '14 at 21:04