This is something of a follow-up to this question that i posted almost a year ago:
It was revealed that two multivariate polynomials $p(x)$ and $q(x)$ have zero sets related to each other by a similarity transformation $A$ if $q(x) = p(Ax)$, where $Ax$ is taken to mean the action of the similarity transformation on the "vector of variables" $x$. In the following, such zero sets will be referenced as having the "same shape". However, that was not my end goal at the time - I was in fact looking to enumerate all the "different shapes" that could be achieved by intersecting the surface represented by the zero set of a given polynomial with hyperplanes spanned by the basis vectors of the space in which the surface is embedded, i. e. the surfaces obtained by setting some of the variables of the polynomial to zero before evaluation. In general, there are $n$ choose $k$ hyperplanes spanned by $k$ basis vectors of an $n$-dimensional vector space. However, intersection of the zero set surface with these can yield the "same shape" in cases.
I did in fact obtain an enumeration procedure, which was verified to yield correct results in dimensions low enough for manual computation, but I was still left with the feeling of not entirely understanding what I had discovered. My reasoning was as follows:
Let $p(x) \in \mathbb{R}[x_1\ldots x_n]$ and $Z(p) \subset \mathbb{R}^n$ be its zero set. Let $\Pi \subset \mathbb{R}[x_1\ldots x_n]$ be the orbit of $p$ under actions of the form $p(x) \mapsto p(Ax)$, where $A$ is orthogonal (scaling was irrelevant for my considerations, and I decided to focus on orthogonal transformations). $\Pi$ is immediately seen to be a homogenous space under orthogonal actions. Intersection of $Z(p)$ with a hyperplane spanned by $k$ basis vectors of $\mathbb{R}^n$ is equivalent to setting all but $k$ variables of $p$ to zero, and so the group that leaves invariant the "shape" of the intersection is $\text{O}(k)\times\text{O}(n-k)$, the first factor acting on the non-zero variables, and the second on the zero variables. Thus identifying all intersections of the "same shape", we can take the quotient $\text{O}(n)/\text{O}(k)\times\text{O}(n-k) = \textbf{Gr}(k,\mathbb{R}^n)$, the Graßmannian of $k$-dimensional subspaces of $\mathbb{R}^n$. The dimension of this Graßmannian then immediately reveals the number of different "shapes" that can be obtained by intersecting $Z(p)$ with hyperplanes spanned by basis vectors, assuming that there are no further symmetries to consider. In my specific case, there was another $\text{U}(n)$ symmetry, so the final number of "shapes" was the dimension of the quotient $\textbf{Gr}(k,\mathbb{R}^n)/\text{U}(n)$. Note that this only seems to work for $1 < k < n-1$, or when there are no one-dimensional subspaces involved. I suspect that this is because $O(1)$ has no parameters.
Now, all of this seems to yield correct results for the cases I was able to manually verify, but is lacking rigor left and right so that the deeper meaning (if there is any) is lost on me. I would however really like to have a formal proof of the connection between the dimension of the Graßmannian and the number of "shapes", and any insight into the matter is greatly appreciated.