Let $G$ be a finite group acting linearly on $\mathbb{A}^n$. Do we expect algebraic vector bundles on $X := \mathbb{A}^n/G$ to be trivial? Here by the quotient I mean the singular scheme, not the stack quotient.

What is known:

(1) For finite abelian groups $G$, $X$ is an affine toric variety, and vector bundles on $X$ are trivial by a Theorem of Gubeladze: https://iopscience.iop.org/article/10.1070/SM1989v063n01ABEH003266. In particular, when $G$ is the trivial group, triviality of vector bundles on $\mathbb{A}^n$ is an older result by Quillen-Suslin.

(2) Line bundles on $X$ are trivial. This can be shown by lifting a line bundle on $X$ to a $G$-line bundle on $\mathbb{A}^n$; then $\mathrm{Pic}^G(\mathbb{A}^n)$ is the group of characters of $G$, and since the linear $G$-action has a fixed point $0$, this character will be trivial, hence coming from a trivial line bundle on $X$.

(2') For vector bundles of higher rank the argument in (2) does not work. This has to do in particular with $G$-actions on $\mathbb{A}^N$ not being linearizable in general: https://link.springer.com/chapter/10.1007%2F978-94-015-8555-2_3

(3) The Grothendieck group of vector bundles is $\mathrm{K}_0(X) = \mathbb{Z}$. We prove it in https://arxiv.org/pdf/1809.10919.pdf, Prop. 2.1 indirectly, using comparison with cdh topology of differential forms.

(3') By homotopy invariance of K-groups in the smooth case, $\mathrm{K}^G_0(\mathbb{A}^n) \simeq \mathrm{K}^G_0(\mathrm{Spec}(k))$ which is the Grothendieck ring of $G$-representations; however this does not seem to help.

Is there any more evidence for/against the triviality of vector bundles on $X$? Is this question mentioned anywhere in the literature?