Suppose $X$ is a singular variety over a field $k$, which admits an action by a finite group $G$. Suppose the quotient $X/G$ is also a variety over $k$. If $Y\_G$ is a resolution of $X/G$, does there exist a variety $Y$ which is a resolution of $X$ such that the action on $X$ could be extended to an action on $Y$ and 


$$Y/G \simeq Y\_G$$ 


Put it in another way, is the resolution of the quotient "the same thing" as the quotient of (equivariant) resolution?



Is the quotient of resolution the same as resolution of the quotient?
As I said in the comments, the answer is no in general. Indeed, let $X = \mathbb{P}^2\times \mathbb{P}^2$ and let $G=\mathbb{Z}/2\mathbb{Z}$ act by permuting the factors. Then $X$ is smooth and $X/G$ is singular. If $(X/G)'\to X/G$ is a resolution of singularities, then $(X/G)' \not \cong X/G$ (clearly).


But, if $G$ acts freely on $X$ (in the scheme-theoretic sense), then $X\to X/G$ is a $G$-torsor (for the etale topology). Let $Y\_G \to X/G$ be a desingularization. Then, the pull-back $Y:= X\times\_{X/G} Y\_G$ of $X\to X/G$ along $Y\_G \to X/G$ is a $G$-torsor over $Y\_G$ and maps $G$-equivariantly to $X$. This means that the answer to your question is positive when $G$ acts freely on $X$.


