Checking smoothness of the components of a highly symmetric scheme via quotient? Setting
Let $I\subseteq\mathbb C[x_0,\ldots,x_n]=:S$ be a homogeneous ideal and $X\subseteq\mathbb P^n$ the scheme defined by $I$. Consider the action of the symmetric group $\mathfrak S_{n+1}$ on $S$ by permuting the variables. Assume that $I$ is invariant under under some subgroup $G\subseteq\mathfrak S_{n+1}$. Assume furthermore that $G$ acts transitively and freely on the irreducible components of $X$. In other words, all irreducible components of $X$ are isomorphic and we have one component for each permutation in $G$. You can obtain something like this by picking any (possibly open) point of $\mathbb P^n$ which has trivial stabilizer in $G$ and taking (the closure of) its $G$-orbit.
Question
I am in such a situation and I want to figure out whether each component of $X$ is nonsingular. I thought it might be a good idea to consider the quotient $X/G$, which is defined as $\operatorname{Proj}(S^G/I^G)$. Here, I denote by $S^G := \{ f\in S \mid G.f=\{f\}\}$ the $G$-invariants in $S$.
My question is whether the following is true:

The irreducible components of $X$ are nonsingular if and only if $X/G$ is nonsingular.

Thoughts so far
My intuition tells me that $X/G$ should be isomorphic to each component of $X$ (in the general case, also including its embedded points). If $X$ is normal, then this is easily true: Restricting the projection $\pi:X\twoheadrightarrow X/G$ to any component of $X$ yields a surjective morphism between normal varieties whose fibers generically contain one element, so this morphism is bijective. Because it maps between normal varieties, it is an isomorphism. I am not sure how to treat this in the general case, though.
I went through the examples $I=(x,yz)$ and $I=(x^2,xz,yx,yz)$ in $\Bbb C[x,y,z]$ with $G$ generated by the transposition of $y$ and $z$ only. It behaves as I expected, but I gained no insights.
 A: Here is an example where $X/G$ is singular and the components of $X$ are smooth.
In $\mathbb{P}^3$ with coordinates $x,y,z,w$, consider the smooth conics
$$\begin{array}{lll}X_1:& x^2-y^2=yw,&z=w\\ X_2:& z^2-w^2=yw,&x=y.\end{array}$$
They are exchanged by the involution $\sigma:(x:y:z:w)\mapsto(z:w:x:y)$ and meet only at the points $(0:0:1:1)$ and $(1:1:0:0)$ which are themselves exchanged by $\sigma$. In particular, $\sigma$ acts freely on $X:=X_1\cup X_2$, so the projection $X\to X/\langle\sigma\rangle$ is étale. Hence $X/\langle\sigma\rangle$ is singular (because $X$ is; in fact $X/\langle\sigma\rangle$ is a rational curve with one node) while $X_1$ and $X_2$ are smooth.
A: *

*You apparently mean the action on the set of components is free, not just faithful.

*For $C$ a component of $X$, the composite map $C \to X \to X//G$ is a finite map, and (by the freeness assumption) birational. From there you need only that $X//G$ is normal to infer that this map is an isomorphism. (You don't need any assumption on $C$ nor do you need smoothness of $X//G$.)
I'm trying to involve the square $$\begin{matrix} 
C \times G &\to& X \\ \downarrow &&\downarrow \\ C &\to& X//G\end{matrix}$$
where down-arrows divide by the $G$-action on the right (as indicated by the notation $X//G$) and right-arrows by the diagonal interior action $(c,g)\sim (ch, h^{-1}g)$. Your questions should be about comparing the two horizontal (birational) maps.
