I asked a version of this question earlier, but it was very imprecise and poorly formatted, so I decided to create a new question.
Suppose we have an ordered set of $n(n-1)/2$ distinct polynomials $P=\{P_1,P_2,...,P_{n(n-1)/2}\}$, each of the same degree, such that the product $\prod_{i=1}^{n(n-1)/2}P_i$ is symmetric with respect to the usual action of $S_n$ on $\mathbb{C}[x_1,x_2,...x_n]$.
I'm trying to show that $\forall P_i,P_j \in P$, there exists an automorphism $\phi$ of $\mathbb{C}[x_1,x_2,...x_n]$ such that $\phi(P_i)=P_j$. In fact, there exist $\phi$'s that form subgroup of the automorphism group of $\mathbb{C}[x_1,x_2,...x_n]$ that is isomorphic to $S_n$, where each $\phi$ corresponds to a permutation of $P$.
For example, for $n=3$, working in $\mathbb{C}[x_1,x_2,x_3]$, if we take our set $P$ to be the polynomials $(x_1+x_2),(x_2+x_3),(x_1+x_3)$, the product $(x_1+x_2)(x_2+x_3)(x_1+x_3)$ is symmetric and there exist automorphisms of $\mathbb{C}[x_1,x_2,x_3]$ that send each one to the other.
This too is vaguely phrased, but I hope I got my point across!
Here is the link to the old question:
Product of $n(n-1)/2$ polynomials of the same degree is symmetric
This conjecture is false. Choose a multiple $d=k n$ of $n$, with $k>1$, such that $n(n-1)/2$ divides $d$ (that is, $d/\binom{n}{2} = 2k/(n-1)$ is an integer $r$.) The monomial $x_1^k\cdots x_n^k$, which has total degree $d$, can be divided up into $n(n-1)/2$ degree-$r$ factors in any number of arbitrary ways, having no invariance properties whatsoever.
For example, if $n=3$ you can take $k=2$, and write $$x_1^2 x_2^2 x_3^2=(x_1^2)(x_2 x_3)(x_2 x_3).$$