Let $G$ be a cyclic group acting linearly on $X := \mathbb{A}^n$. If we assume that the quotient $Y:=X/G$ is non-singular, does it follow that $Y \simeq \mathbb{A}^n$? If so, is it necessary to assume that $G$ is cyclic or that the action is linear?
Context: I'm reading Weighted Projective Varieties by I. Dolgachev. In his proof that if $\mathbb{P}$ is a nonsingular weighted projective space, then it is isomorphic to ordinary projective space, he identifies weighted projective space locally with the quotient of $\mathbb{A^n}$ by a diagonal action of a cyclic group, then shows that if the invariant algebra of this action is isomorphic to a graded polynomial ring (i.e. If the quotient is isomorphic to affine space), then the weights must be the ordinary weights.
The Chevalley–Shephard–Todd theorem implies that if a finite group acting $G$ linearly (and faithfully, w.l.o.g) on $A^n$ has smooth quotient $A^n/G$, then $G$ is generated by pseudoreflections and the quotient is $A^n$.
If we know further that $G$ is cyclic, then there is a generator $g$ of $G$ which is a pseudoreflection. Up to linear change of coordinates, we can therefore assume that the action of $g$ is given simply by $g(x_1,\dots,x_n)=(\lambda x_1,x_2,\dots,x_n)$ with $\lambda$ an $|G|$th root of unity.
You can therefore compute the invariants and see what $A^n/G$ is! :-)
It is not necessary for the group to be cyclic: the standard example is $S_n$ acting on $A^n$ by permutation of coordinates — the quotient in this case is $A^n$.