So, I feel like I'm missing something obvious, but I have the following situation:


Let $X\to Y$ be a finite group quotient of schemes (in fact, varieties) by the finite group $G$. Let $\tilde{Y}\to Y$ be any resolution of singularities. Then we have a natural map $X\times\_Y \tilde{Y}\to X$ which is birational, and $X\times\_Y \tilde{Y}\to \tilde{Y}$ is a quotient by $G$. I want to be able to say that $X\times\_Y \tilde{Y}$ is a resolution of $X$, and if $G$ acts freely, I can see how. But what if there's a locus (let's say a divisor) where $G$'s action has stabilizers?


EDIT: I'm leaving the above, though that situation is no good, thanks to comments below and abx's answer. But perhaps this is more reasonable:


Given $X\to Y$ a quotient by $G$ a finite group and $\tilde{Y}\to Y$ a resolution of $Y$, can I always find a resolution $\tilde{X}\to X$ such that $G$ acts on $\tilde{X}$ and $\tilde{X}/G\cong \tilde{Y}$?



Finite Quotients and Resolutions of Singularities
I suspect that you will have trouble with trying to "simultaneously" resolve $X$ and $X/G$. We actually run into this issue in the joint paper with Anatoly Libgober arXiv:math/0206241 (although it may not be evident from the paper) and consequently had to settle for working with $\hat X\to\hat Y$ which was $G$-equivariant morphism of smooth varieties which is not necessarily a quotient map. It didn't matter for us a whole lot, since we were working with the formalism of log-terminal pairs. 


However, it may not be completely impossible to find this "simultaneous resolution", especially if you know something about $X$ (say it is of small dimension). You can start by using abelianization trick of Batyrev (arXiv:math.AG/0009043) to reduce to the situation when the stabilizers of the action of $G$ are abelian. But then you get into a bit of trouble.


Let me illustrate this trouble by the simplest example, that of $A\_2$ singularity. 
Let $G=\mathbb Z\_3$ act on $X=\mathbb C^2$ by $(x,y)\mapsto (xw,yw^{-1})$ where $w=\exp(2\pi I/3)$. 


One may view the quotient in terms of toric geometry: one starts with the toric variety that corresponds to the cone $C=(\mathbb Q\_{\geq 0})^2$ in the lattice $N=\mathbb Z^2$. Taking the quotient by the above group has the effect of enlarging the lattice $N$ to a coindex three suplattice $N\_1=\mathbb Z^2+ {\mathbb Z}(\frac 13,-\frac 13)$. The minimal resolution of singularities of the quotient corresponds to subdividing the cone $C$ into three cones 
$$
\mathbb Q\_{\geq 0}(1,0)+\mathbb Q\_{\geq 0}(\frac 23,\frac 13),~~
\mathbb Q\_{\geq 0}(\frac 23,\frac 13)+\mathbb Q\_{\geq 0}(\frac 13,\frac 23),~~
\mathbb Q\_{\geq 0}(\frac 13,\frac 23)+\mathbb Q\_{\geq 0}(0,1)
$$
which are generated by lattice elements of $N\_1$ (the two additional rays correspond to the two exceptional divisors of the minimal resolution).


Taking the obvious $\hat X$, which is the normalization of resolution of $Y$ in the field of functions of $X$ corresponds to considering the same subdivision, but now in the original lattice $N$. The same cones can be written now in terms of minimum generators as
$$
\mathbb Q\_{\geq 0}(1,0)+\mathbb Q\_{\geq 0}(2,1),~~
\mathbb Q\_{\geq 0}(2,1)+\mathbb Q\_{\geq 0}(1,2),~~
\mathbb Q\_{\geq 0}(1,2)+\mathbb Q\_{\geq 0}(0,1).
$$
The middle cone is now not generated by a basis of the lattice (determinant is three) and thus gives a singular point.


If one is looking for a toric "simultaneous" resolution of this singularity, then one needs to make a subdivision in such a way that for each of the smaller cones taking minimum generators in either lattice $N$ or $N\_1$ gives a basis of the said lattice. This requires that in each cone exactly one of minimum generators of $N$ coincides with that of $N\_1$.
It is possible to do it in this example by splitting the middle cone into 
$$
\mathbb Q\_{\geq 0}(2,1)+\mathbb Q\_{\geq 0}(1,1),~~
\mathbb Q\_{\geq 0}(1,1)+\mathbb Q\_{\geq 0}(1,2)
$$
which becomes 
$$
\mathbb Q\_{\geq 0}(\frac 23,\frac 13)+\mathbb Q\_{\geq 0}(1,1),~~
\mathbb Q\_{\geq 0}( 1,1)+\mathbb Q\_{\geq 0}(\frac 13,\frac 23)
$$
in $N\_1$.


So in this example such $\hat X$ exists. You can try playing with other surface singularities, say $A\_n$ for starters. If I were to bet on this, I would say that if you can handle $A\_4$, then you should be able to get all of $A\_n$.


