$\DeclareMathOperator{\SL}{\operatorname{SL}}$Let $\mathcal{P}\_{n-1}$ be the space of complex polynomials in one variable, say $z$, of degree at most $n-1$. As a complex vector space, it is clearly $n$-dimensional. Consider the basis $1$, $z,\ldots,z^{n-1}$ of $\mathcal{P}\_{n-1}$. This allows us to identify $\mathcal{P}\_{n-1}$ with $\mathbb{C}^n$.


The complex Lie group $\SL(n,\mathbb{C})$ acts on $\mathbb{C}^n$, and thus, via our identification, it acts on $\mathcal{P}\_{n-1}$. The discriminant is a homogeneous polynomial on $\mathcal{P}\_{n-1}$ of degree $2n-2$.


My question can now be formulated. What is the subgroup of $\SL(n,\mathbb{C})$ which preserves the discriminant (using our identification of $\mathcal{P}\_{n-1}$ with $\mathbb{C}^n$)? I have a feeling it is the image of $\SL(2,\mathbb{C})$ under a principal homomorphism from $\SL(2,\mathbb{C})$ to $\SL(n,\mathbb{C})$.


More specifically, $\SL(2,\mathbb{C})$ acts simultaneously on all linear factors of a polynomial of degree $n-1$, and this action preserves the discriminant. I think these are all "projective" transformations which preserve the discriminant, but I am not sure how to show that. Edit: I should really write that I think these are all transformations in $\SL(n,\mathbb{C})$ which preserve the discriminant (see the remark by @NoamD.Elkies below).


Edit 2: here is a conceptual proof for polynomials of degree at most $2$ (i.e. for $n=3$ using my notation). Let $L$ be the map on $\mathbb{C}P^2$ induced by some element $g \in \SL(3,\mathbb{C})$ which preserves the discriminant of quadratic polynomials. Note that we can think of $\mathbb{C}P^2$ as the set of unordered pairs of points on $\mathbb{C}P^1$ (namely the roots of the corresponding polynomial, viewed up to scaling). Note also that polynomials of vanishing discriminant correspond to an unordered pair of coinciding points on $\mathbb{C}P^1$.


Let $f$ be the holomorphic map from $\mathbb{C}P^1$ to $\mathbb{C}P^1$ obtained essentially by restricting $L$ to coinciding pairs of points on $\mathbb{C}P^1$, in turn corresponding to the vanishing locus of the discriminant. Note that $f$ induces a holomorphic map $\tilde{f}$ on the set of unordered pairs of points on $\mathbb{C}P^1$, and thus on $\mathbb{C}P^2$.


Moreover, $\tilde{f}$ and $L$ agree on the conic in $\mathbb{C}P^1$ coinciding to the vanishing locus of the discriminant, and thus they must be equal, since they are two linear isomorphism maps from $\mathbb{C}P^2$ to itself which agree on a conic in $\mathbb{C}P^2$.


We have thus shown the claim for $n=3$. I am hoping this can be generalized to higher $n$'s. I think it is doable.


Edit 3: the previous argument can be generalized using an additional hypothesis, that the element $g \in \SL(n,\mathbb{C})$ not only preserves the discriminant, but also maps any polynomial with a single root of multiplicity $n-1$ to a polynomial with a single root of multiplicity $n-1$. Under this additional hypothesis, essentially the same argument proves that $L$, which is the map on $\mathbb{C}P^{n-1}$ induced by $g$, is induced by some holomorphic automorphism of $\mathbb{C}P^1$, and thus by some element of $\SL(2,\mathbb{C})$. But is it necessary to make this assumption? Is my conclusion false without it? Or is it perhaps the case that this hypothesis can be proved?



What is the subgroup of $\mathrm{SL}(n,\mathbb{C})$ which preserves the discriminant?
At the request of the OP, I post my comment as an answer. View $\mathcal{P}\_n$ as the space of homogeneous polynomials of degree $n$ in 2 variables. Let $\Delta \_p\subset \mathcal{P}\_n$ be the locus of polynomials with one linear factor of multiplicity $\geq p$. One can show that
the singular locus of $\Delta \_p$ is $\Delta \_{p+1}$. Therefore the subgroup $G$ of $\operatorname{GL}(\mathcal{P}\_n) $ preserving the discriminant hypersurface $\Delta \_2$ preserves $\Delta \_n$. Now the image of $\Delta \_n$ in $\mathbb{P}(\mathcal{P}\_n)\cong \mathbb{P}^n$ is a rational normal curve, that is, the image of the $n$-th Veronese embedding $V\_n:\mathbb{P}^1\hookrightarrow \mathbb{P}^n$. Thus up to homotheties, $G$ is the group of automorphisms of $\mathbb{P}^n$ preserving
$V\_n(\mathbb{P}^1)$, which maps isomorphically to $\operatorname{Aut}(\mathbb{P}^1)=\operatorname{PGL}(2,\mathbb{C}) $.


