$\mathbb{C}[x_1, \dots, x_n]^G \cong \mathbb{C}[x_1, \dots, x_n]^H$ with $G$ and $H$ small implies $G \cong H$? We say that $g \in \text{GL}(n,\mathbb{C})$ is a reflection if it fixes a hyperplane, and say that a finite subgroup $G$ of $\text{GL}(n,\mathbb{C})$ is small if it contains no reflections. My question is then as follows:

Suppose that $G$ and $H$ are small subgroups of $\text{GL}(n,\mathbb{C})$, and let $G$ and $H$ act naturally on the polynomial ring $R = \mathbb{C}[x_1, \dots, x_n]$. If $R^G \cong R^H$, does it follow that $G \cong H$?

Without the smallness hypothesis, this is massively false. Unless I'm mistaken, it's true in dimension 2, but the only reason I know this is by appealing to the classification of small subgroups of $\text{GL}(2,\mathbb{C})$.
 A: $\def\CC{\mathbb{C}}\def\Spec{\mathrm{Spec}\ }$ This answer assumes you know some algebraic geometry; I don't know how you would do this by pure algebra. Recall that the $\CC$-valued points of $\Spec R$, with the analytic topology, are $\CC^n$ with its standard topology. Likewise, the $\CC$-valued points of $\Spec R^G$, with the analytic topology, are the quotient space $\CC^n/G$ with its quotient topology. So the point is to show that we can recover the group $G$ from the space $\CC^n/G$.
I'll write $f$ for the map $\CC^n \to \CC^n/G$.
Let $X$ be the set of points in $\CC^n$ with trivial stabilizer. Clearly, $X \to X/G$ is a covering map, with Deck group $G$. Also, all non-trivial subgroups of $G$ fix a subspace of codimension $\geq 2$ (by the small condition), so $X$ is $\CC^n$ with a finite list of codimension $\geq 2$ subspaces removed. This means that $X$ is simply connected. Thus, $\pi_1(X/G) \cong G$. Thus, we just need to show that we can intrinsically recognize the subset $X/G$ of $\CC^n/G$.
Lemma: Let $x \in \CC^n$. Then $x \in X$ if and only if $f(x)$ is a smooth point of $\CC^n/G$.
Proof: The standard criterion, for an arbitrary $G$-action, is that $f(x)$ is smooth if and only if the stabilizer of $x$ is generated by reflections. Since your $G$ is small, the only subgroup generated by reflections is the trivial subgroup. $\square$.
So, putting it all together: We can recover the space $\CC^n/G$ as the $\CC$-valued points of $\Spec R^G$, with the analytic topology. We can recover $X/G$ as the smooth locus in $\CC^n/G$. Then $G \cong \pi_1(X/G)$.
