Problem. Let $G\times X\rightarrow X$ be a continuous action of a Polish group on a Polish space. Let $A\subseteq X$ be Borel. Is the saturation $[A]_{G}:=G\cdot A$ a Borel set?
One approach. The collection $\{A\in \text{Bor}(X):G\cdot A\in \text{Bor}(X)\}$ obviously contains all open sets and is closed under countable unions. The only problem seems to be that it not be closed under complements.
Another approach. The Vaught-transformations $A^{\Delta}:=\{x\in X:(\exists^{\ast}{g\in G})~gx\in A\}$ and $A^{\ast}:=\{x\in X:(\forall^{\ast}{g\in G})~gx\in A\}$ transform Borel sets into Borel sets. The following equation holds for saturation:
$$A^{\Delta}\subseteq [A]_{G}$$
But this does not help much.
Question. Is there another approach? Or are there only some kinds of actions which yield that saturations of Borel sets are Borel?
unless I am overlooking something, I think you could use projection in place of saturation showing the answer is no.
Let $G$ be the real line and $X$ be the plane, with the action defined as $(g,(x,y))\to(x,g+ y)$.
Take any $A\subset X$ that is Borel, but whose projection to the $x$-axis is not Borel (it would only be analytic). The projection of $A$ to the $x$-axis would be the same as the saturation of $A$ (in the plane) intersected with the $x$-axis. If the saturation was Borel, then the projection (being the intersection of two Borel sets) would also be Borel. But since the projection of $A$ is not Borel we conclude that the saturation of $A$ is not Borel either.