Let $I\subseteq R:=\mathbb C[x_0,\ldots,x_n]$ be a homogeneous ideal defining a subscheme $X\subseteq\Bbb P^n$. As in my previous question, the permutation group $\mathfrak S_{n+1}$ acts on $R$ by permuting the variables, inducing an action on $\Bbb P^n$. and there is a subgroup $G\subseteq\mathfrak S_{n+1}$ which leaves $X$ invariant. I would like to (practically) compute the dimension of the Zariski Tangent space of $Y:=X/\!/G$ at a point $y$. Here, $Y$ is defined as $\operatorname{Proj}(R^G/I^G)$.

Let $\pi\colon X\twoheadrightarrow X/\!/G = Y$ be the projection and $\pi(x)=y$. Computing the tangent space of $X$ at some point $x$ is rather *"easy"* in the sense that I can do it practically: I can (in a computer algebra system) provide generators for $I$ and then compute the kernel of the Jacobian. However, I am not sure how to compute the tangent space of $Y$ at $y$ efficiently.

There are computer algebra methods to compute generators of the invariant ring $R^G$ and also generators for $I^G$ in terms of these, but I have tried them and my examples seem too large for them to handle, even though $R^G$ is a polynomial ring by the Chevalley–Shephard–Todd Theorem.

Maybe I am doing something wrong, but mostly I think it might be easier to compute just the tangent space. Note however that the differential $d\pi\colon \mathcal T_{X,x}\to \mathcal T_{Y,y}$ does not have to be surjective: Given a (homogeneous) prime ideal $\mathfrak p\subseteq R/I$, we only have $(\mathfrak p^G)^2\subseteq(\mathfrak p^2)^G$, usually equality will not hold, so $\mathfrak p^G / (\mathfrak p^G)^2 \to \mathfrak p / \mathfrak p^2$ is not generally injective. Even if something like this could be said however, I do not know what the map $d\pi$ even looks like.

The smallest case that interests me is $n=36$, where $\Bbb P^n$ can be though of as (classes of) $6\times 6$ matrices and $G\cong\mathfrak S_3$ acts by permuting the first $3$ rows of a matrix. I will not go into detail about $X$ unless requested, I guess it is enough to know that it is given by explicit equations and computing a Groebner basis for the ideal $I$ is presumably out of reach.