Let $X\in\mathbb{R}^{d\times d}$ be the diagonal matrix with $d/2$ entries equal to $1$ and $d/2$ entries equal to $-1$. Let $F_U \triangleq \frac{1}{d}\|\operatorname{diag}(U^{\dagger}XU)\|^2_F$ denote the sum of squares of the diagonal entries of $U^{\dagger}XU$. If $U$'s columns consisted of independent random complex unit vectors, then we know that $\mathbb{E}[F_U] = \frac{1}{d+1}$, and $\Pr\left[\left|F_U - \frac{1}{d+1}\right| > t \right] \le e^{-\Omega(t/d^{3/2})}$.
Now suppose $U$ is Haar-random unitary, in which case we still have that $\mathbb{E}[F_U] = \frac{1}{d+1}$. My question is: does $F_U$ concentrate comparably in this case?
Weingarten calculus tells us that $$\mathbb{E}[F_U^n] = \sum_{\substack{\sigma,\tau\in \mathbb{S}_{2n}: \\ \tau \text{ has only even cycles}}} d^{\kappa(\tau)}\cdot \operatorname{Wg}(\sigma\tau^{-1},d)\cdot \Pr_{s\in[d]^n}[\sigma \text{ fixes the word } (s_1,s_1,s_2,s_2,\ldots,s_n,s_n)],$$ where $\kappa(\cdot)$ denotes number of cycles, but it is not clear to me how to get a good estimate for this quantity for large $n$. Alternatively, perhaps a stretch, but is there some HCIZ-esque way of computing the MGF of $F_U$?
