$\newcommand{\pt}{\mathit{pt}}$For $d>1$ is it possible to understand $\text{Aut}(\text{Sym}^n(\mathbb{P}^d\_{\mathbb{C}}))$? Automorphisms mean biregular morphisms from the variety to itself. Not let $\pt$ be a point on the projective space. Consider the point on $\text{Sym}^n(\mathbb{P}^d\_{\mathbb{C}})$ corresponding to the point $\pt$ but with degree $n$. Let's denote it by $\pt^n$. What is the orbit of $\pt^n$ under $\text{Aut}(\text{Sym}^n(\mathbb{P}^d\_{\mathbb{C}}))$? Do they necessarily get mapped to another point of the form $\pt\_2^n$?



Automorphisms of symmetric powers of projective space
$\DeclareMathOperator\PGL{PGL}$Yes, it's possible to understand this automorphism group, and yes, such a point is necessarily mapped to another point of the same form.


I claim the automorphism group is given by $\PGL\_{d+1}$, acting in the way induced by its action on $\mathbb P^d$. The statement about orbits follows immediately.


To prove this, let's consider the smooth locus of $\operatorname{Sym}^n(\mathbb P^d)$. I claim it consists of only distinct unordered tuples of points. Since the singular locus is closed, it suffices to check that the point $(x\_1,x\_1, x\_2,\dots, x\_{n-1})$ is not smooth, for which it suffices to check that its tangent space has dimension $d (n-2) + d+ d (d+1)/2> nd$.


Now the map $(\mathbb P^d)^n \to \operatorname{Sym}^n(\mathbb P^d)$, restricted to the smooth locus, is a finite etale covering from $(\mathbb P^d)^n$ minus the big diagonal to the smooth locus. Since $\mathbb P^d$ is simply-connected, and the big diagonal has codimension $d >1$, the complement of the big diagonal is simply-connected and thus equal to the universal cover of the smooth locus.


Now every automorphism of $\operatorname{Sym}^n(\mathbb P^d)$ restricts to an automorphism of the smooth locus, hence lifts to an automorphism of its universal cover, thus extends to an automorphism of the normalization of the variety in that cover, which is $(\mathbb P^d)^n$. The automorphism group of $(\mathbb P^d)^n$ is the semidirect product $(\PGL\_{d+1})^n \rtimes S\_n$, so it must arise from one of those, but only the subgroup $\PGL\_{d+1} \times S\_n$ preserve the map to $\operatorname{Sym}^n(\mathbb P^d)$ and give automorphisms of $\operatorname{Sym}^n(\mathbb P^d)$. Since $S\_n$ gives trivial automorphisms of $\operatorname{Sym}^n(\mathbb P^d)$, every automorphism must come from $\PGL\_{d+1}$.


