Let $\{e\_1,\ldots, e\_n\}$ be the standard basis of $\mathbb{C}^n$. Consider the $m$-multilinear form $$v=\sum\_{i=1}^n e\_i^{\otimes m}\in (\mathbb{C}^n)^{\otimes m}$$
and consider the action of $\text{GL}\_n(\mathbb{C})$ on
$(\mathbb{C}^n)^{\otimes m}$.


Question: What is the stabilizer of $v$ in $\text{GL}\_n(\mathbb{C})$ where $m>2$?
If $m=2$ this is just the orthogonal group. If $m>2$ then this must contain the subgroup $S\_n\wr C\_m$, where the $i$-th copy of $C\_m$ acts nontrivialy only on $e\_i$. Does the stabilizer equal to this wreath product, or can it be bigger?



Stabilizers of multilinear forms
It is equal.


Since the ground field is of characteristic zero and $v$ is symmetric one may as well compute the stabilizer of the polynomial $f:=\sum\_{i=1}^nx\_i^m$.


Assume $g\in\mathrm{GL}\_n(\mathbb C)$ stabilizes $f$. Then it also stabilizes the Hessian $$\det\nolimits\_{ij}(\partial\_i\partial\_jf)\in\mathbb C^\*(x\_1\cdots x\_n)^{m-2}$$ up to a factor. Hence it permutes the irreducibel factors $x\_1,\ldots,x\_n$ up to scalars, i.e., $g\in\mathbb C^\*\wr S\_n$. Plugging this back into $f$ we conclude $g\in C\_m\wr S\_n$.


