Let $G$ be a finite algebraic group acting on a projective complex variety $X$. Then a quotient $Y=X/G$ exists as a scheme and, if $G$ acts freely, $Y$ is an orbit space and the natural map $$\eta:[X/G]\to X/G=Y$$ is an isomorphism. I am trying to figure out how this isomorphism works at the level of points.
The $\mathbb C$-points of $Y$ are the orbits of the action. These must correspond to the $\mathbb C$-points of the stack $[X/G]$, which are $G$-equivariant maps $G\to X$. I am trying to see how this correspondence works. I would say that a $G$-orbit $\textrm{Orb}(x)\subset X$ should correspond to the $G$-equivariant map $\alpha_x:G\to X$ sending $g\mapsto g\cdot x$. The orbit is just the image of this map. But conversely, if I have a $G$-invariant map $G\to X$, is it obvious that its image is an orbit?
Let now $S$ be any scheme over $\mathbb C$. The $S$-points of the stack $[X/G]$ are couples $(P,f)$ where $P\to S$ is a principal $G$-bundle and $f:P\to X$ is a $G$-equivariant map. I cannot describe explicitly the $S$-points of $Y=X/G$.
Question. Is it true, at least étale locally, that the $S$ points of $Y$ are the orbits of the action of $G(S)$ on $X(S)$? How do the $S$-points of $Y$ compare to those of $[X/G]$?
Thanks!