This has been asked on MSE here, but has not had much traffic so I will ask a similar question here as many people here enjoy topics around resolving singularities.

Let $X$ be a complex threefold with an action by a finite group $G$. Suppose the quotient $X/G$ is singular, and admits a crepant resolution, for example $G\subset \operatorname{SL}_3(\mathbb{C})$ by Bridgeland-King-Reid. Instead of working with the quotient, can you blowup the fixed locus on $X$ to get fixed divisors so that the quotient of the blowup is smooth, and a crepant resolution of the quotient? In other words, is there some $\widetilde{X}$ such that the following diagram commutes \begin{array}{c} \widetilde{X} & \to & \widetilde{X/G} & \\ \downarrow& & \downarrow \\ X & \to & X/G \end{array} where $\widetilde{X/G}$ is a crepant resolution of $X/G$?

This is possible with transversal $A_n$ singularities, and triple Veronese singularities, but for example if $G = \langle \zeta_6\times \zeta_3\times \zeta_2\rangle$ acts on $\mathbb{C}^3$, where $\zeta_n$ is a primitive $n$-th root of unity, then the $y$-axis is a transveral $A_1$, the $z$-axis is a transversal $A_2$, and these can be blown up to fixed divisors, but the origin fixed by all of $G$, the intersection of the two transversal axes, is more difficult. I have tried different ways and every approach just gives extra points fixed by $G$ that I have not been able to get a corresponding divisor.

My motivation for asking is that I am hoping to avoid having to work with the singularities in $X/G$ (which are quite nasty for some of the groups I am interested in) and instead just work with affines instead, but I have been blowing up and contracting in many different ways an unable to find a desired resolution. After many attempts, I realized it may not be possible, though perhaps I simply need to get more creative with my approach.

This has been asked on MSE here, but has not had much traffic so I will ask a similar question here as many people here enjoy topics around resolving singularities.

Let $X$ be a complex threefold with an action by a finite group $G$. Suppose the quotient $X/G$ is singular, and admits a crepant resolution, for example $G\subset \operatorname{SL}_3(\mathbb{C})$ by Bridgeland-King-Reid. Instead of working with the quotient, can you blowup the fixed locus on $X$ to get fixed divisors so that the quotient of the blowup is smooth, and a crepant resolution of the quotient? In other words, is there some $\widetilde{X}$ such that the following diagram commutes \begin{array}{c} \widetilde{X} & \to & \widetilde{X/G} & \\ \downarrow& & \downarrow \\ X & \to & X/G \end{array} where $\widetilde{X/G}$ is a crepant resolution of $X/G$?

This is possible with transversal $A_n$ singularities, and triple Veronese singularities, but for example if $G = \langle \zeta_6\times \zeta_3\times \zeta_2\rangle$ acts on $\mathbb{C}^3$, where $\zeta_n$ is a primitive $n$-th root of unity, then the $y$-axis is a transveral $A_1$, the $z$-axis is a transversal $A_2$, and these can be blown up to fixed divisors, but the origin fixed by all of $G$, the intersection of the two transversal axes, is more difficult. I have tried different ways and every approach just gives extra points fixed by $G$ that I have not been able to get a corresponding divisor.

My motivation for asking is that I am hoping to avoid having to work with the singularities in $X/G$ (which are quite nasty for some of the groups I am interested in) and instead just work with affines instead, but I have been blowing up and contracting in many different ways an unable to find a desired resolution. After many attempts, I realized it may not be possible, though perhaps I simply need to get more creative with my approach.