Context: Inspired by a recent question I found a very cute proof that the zero set of a complex polynomial depends continuously on the polynomial. (Yes, there's some question regarding what that even means; will clarify that if and when I talk about it - this post is about something else.) Just realized there's a problem with the argument - this question is about fixing that.
Say $p_0,\dots p_d$ are the elementary symmetric polynomials in $d$ variables. We all know that if $p\in\Bbb C[z_1,\dots,z_d]$ is symmetric then there exists a polynomial P such that $$p=P(p_0,\dots,p_d).$$
So far these polynomials are what I think of as holomorphic polynomials on $\Bbb C^d$. Now suppose instead that $p$ is a "real polynomial" on $\Bbb C^d$: $$p=p(z_1,\dots,z_d;\overline z_1,\dots,\overline z_d).$$And suppose $p$ is symmetric, in the sense that $$p(z_1,\dots,z_d;\overline z_1,\dots,\overline z_d)=p(z_{\sigma(1)},\dots,z_{\sigma(d)};\overline z_{\sigma(1)},\dots,\overline z_{\sigma(d)})\quad(\sigma\in S_d).$$
Question: Does it follow that there exists a polynomial $P$ such that $$p=P(p_0,\dots,p_d;\overline p_0,\dots,\overline p_d)?$$
Note: Of course we could just as well say $$p=p(x_1,\dots,x_d;y_1,\dots,y_d).$$But $p$ is not invariant under an arbitrary permutation of $(x_j,y_j)$, just under the subgroup of $S_{2d}$ that maps $(x_j)$ to itself and then applies the same permutation to $(y_j)$.
(My work so far: No clue. Maybe it's true; I can imagine some sort of "algebraic" Stone-Weierstrass on $\Bbb C^2/S_d$. Otoh if it's true it should be clear for, say, $d=2$ and $p(z)=|z|^2$; I get as far as $$|z|^2=(z_1+z_2)(\overline{z_1+z_2})-(z_1\overline z_2+\overline z_1z_2)$$and I'm stuck. If it's false I bet that's a counterexample...)