Let $X$ be an affine irreducible variety, $\operatorname{Birat}(X)$ the group of birational automorphisms of $X$ onto itself, and $G$ a finite subgroup of $\operatorname{Birat}(X)$.
Is there an affine open subvariety $U$ of $X$ such that $U$ is $G$-invariant and each $g \in G$ restricted to $U$ becomes a regular map on $U$?
I firmly believe that the answer is YES and this is easy to show, but somehow I an unable to come up with an argument.
For each $g\in G$ there is an open subset $U_g\subset X$ such that $g\colon U_g\to g(U_g)$ is an isomorphism.
Now $U:=\bigcap_{g\in G}U_g$ is such that for each $g\in G$, the restriction $g\colon U\to g(U)$ is an isomorphism. However, $U$ is not necessarily $G$-invariant yet.
Letting $\tilde U:=\bigcap_{g\in G}g(U)$, we have that $\tilde U$ is $G$-invariant, and that the restriction $g\colon \tilde U\to\tilde U$ is an isomorphism. (In other words, $\tilde U:=\bigcap_{g,h\in G}g(U_h)$.)