TITLE: Total sum of squares of characters of the symmetric group $\mathfrak{S}_n$
QUESTION [7 upvotes]: In my earlier MO post, I proposed the double sum $\sum_{\mu\vdash n}\sum_{\lambda\vdash n}\chi_{\mu}^{\lambda}$ regarding characters of the symmetric group $\mathfrak{S}_n$. Soon after, I started considering the sum of squares $\sum_{\mu\vdash n}\sum_{\lambda\vdash n}(\chi_{\mu}^{\lambda})^2$ hoping to gain a better formula. A further look into older MO posts here and also here shows a Burnside-type Lemma
$$\frac{1}{n!} \sum_{\alpha \in \frak{S}_n} \left( \sum_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$
After comparing the last two sums, I got prompted to ask:

QUESTION. The numerics suggest the below equality. Why is this true?
$$\sum_{\mu\vdash n}\sum_{\lambda\vdash n}(\chi_{\mu}^{\lambda})^2
=\frac{1}{n!} \sum_{\alpha \in \frak{S}_n} \left( \sum_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$

REPLY [3 votes]: I think this is true. Let $X$ be the character table of $S_{n}$, viewed just as a $p(n) \times p(n)$ matrix. Then the left side is ${\rm trace}(XX^{T}) = {\rm trace}(X^{T}X).$ This is $\sum_{\alpha} |C_{S_{n}}(\alpha)|$, where $\alpha$ runs over a set conjugacy class representatives for $S_{n}.$ As mentioned in my comment, the right side is clearly $\frac{1}{n!} \left( \sum_{\alpha} |S_{n}| |C_{S_{n}}(\alpha)| \right),$ where $\alpha$ again runs through class representatives for $S_{n},$ so both sides are equal.
Later edit: For this result, the only special property of $S_{n}$ used is that its complex irreducible characters are all real valued. The following similar equality is true for a general finite group $G$ by a slight modification of the argument above (a straightforward consequence of the orthogonality relations).
$$\sum_{(\chi,x)}|\chi(x)|^{2} = \frac{1}{|G|}\sum_{y \in G} \left( \sum_{\chi} |\chi(y)|^{2} \right)^{2},$$
where $\chi$ runs through the distinct irreducible complex characters of $G$ and $x$ runs over a set of representatives for the conjugacy classes of $G$.