Thanks for rephrasing it in this language. I'll browse SGA for statement like this tomorrow.

Thanks. This proof was very close to my original proof. I felt that my extra assumptions had more to do with personal comfort than actual mathematics.

Thanks for the comment. I am aware what the answer should be. As in my example, $k[G] \cong \oplus_{\overline{g} \in G/H} g k[H]$. We want to pick out one of any of these copies $k[H]$ (with $H$ action), all of which are isomorphic under the action of $G/H$. There should be descent data describing how to pass to a sheaf on the point. If $H=\{e\}$ then it is just taking $G$-invariants. It should be something like taking $G/H$-invariants -- but this doesn't make any sense on most $G$ modules (e.g. $k[G]$). I am curious how the descent data would encode this interesting example.