Let $Y$ be a scheme and a finite group $G$ acts on $Y$ in the sense that $G$ embeds in $Aut_{Sch}(Y)$ in category of schemes. I have two question concerning general properties dealing with action by finite groups on schemes.
Let $T$ any other scheme and we consider the set $Hom(T, Y)$. Clearly the action on $Y$ by $G$ induces naturally action on the set $Hom(T, Y)$ by $G$.
Question #1: Can we reconstruct the action on $G$ if we know completly how $G$ acts on sets $Hom(T, Y)$ for all schemes $T$?
Now it's natural to ask when we can build the 'quotient' $Y/G$ and when does it live also in category of schemes. The 'quotient' $Y/G$ satisfies following properties:
i) $p: Y \to Y/G$ is $G$ invariant
ii) every $G$-invariant morphism $Y \to T$ factors over $Y/G$.
I know that this always the case if there exist open afine cover $Y= \cup V_i$ where every $V_i= Spec(R_i)$ is $G$-invariant in the sense that the $G$-action by automorphisms on $Y$ restricts for every $i$ to automorphism on $V_i$. Then $G$ acts on every ring $R_i$ and we can build the invariant rings $R_i^G \subset R_i$. It $G$ is finite and $R_i$ 'nice' then $R_i^G \subset R_i$ is finite. Then we can define $Y/G$ by glueing the affine schemes $U_i= Spec(R_i^G)$ so $Y/G$ is a scheme.
Question #2: Can we here go 'backwards'? So assume that we know that the quotient $Y/G$ is a scheme, can we find an afine cover $Y= \cup V_i$ where every $V_i= Spec(R_i)$ is $G$-invariant.