This sounds like something that must have been answered long ago, but for some reason I can find nothing on it in the internet. (There has been lots of recent activity in diagonal covariants, related to the $n!$ conjecture, but invariants seem to have become a stepchild in this process.)
Let $k$ be a field of characteristic $0$. Let $n\in\mathbb N$. The group $S_n\times S_n$ acts on the polynomial ring $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]$ by
$\left(\sigma,\tau\right)\left(P\right) = P\left(X_{\sigma^{-1}\left(1\right)},X_{\sigma^{-1}\left(2\right)},...,X_{\sigma^{-1}\left(n\right)},Y_{\tau^{-1}\left(1\right)},Y_{\tau^{-1}\left(2\right)},...,Y_{\tau^{-1}\left(n\right)}\right)$P\left(X_{\sigma\left(1\right)},X_{\sigma\left(2\right)},...,X_{\sigma\left(n\right)},Y_{\tau\left(1\right)},Y_{\tau\left(2\right)},...,Y_{\tau\left(n\right)}\right)$.
Thus, the symmetric group $S_n$ also acts on $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]$ due to the diagonal embedding $S_n\to S_n\times S_n$.
Since the action of $S_n$ is not generated by pseudoreflections, it follows from the converse of the Chevalley-Shephard-Todd theorem that $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]$ is not a free $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]^{S_n}$-module. But it is easy to see that $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]$ is a free $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]^{S_n\times S_n}$-module of rank $n!^2$.
Question: Is $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]^{S_n}$ a free $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right] ^ {S_n\times S_n}$-module?
