Consider the following element $A$ in $U(n)$: $$ \begin{pmatrix} 1/2(1+z) & 1/2(1-z) & \\\\ 1/2(1-z) & 1/2(1+z) & \\\\ & &I_{n-2} \end{pmatrix},$$ where $|z| = 1$.

Now conjugate $A$ by permutation matrices $S$, i.e., $S(i,j) =1$ if $\sigma(i) = j$ for a fixed $\sigma \in S_n$. What group does $S A S^{-1}$ generate? What is its dimension?

Finally let $A' = \begin{pmatrix} i & -i & \\\\ -i & i & \\\\ & & 0_{n-2} \end{pmatrix}$ be the associated Lie algebra element to $A$ (i.e., derivative with respect to $z$ at $z = 1$; notice the condition $|z|=1$). Can one give an explicit basis of the Lie algebra closure generated by $S A S^{-1}$, with each element $B$ in the basis of the form $Ad(Ad(\ldots Ad(Ad(Ad(Ad(S_1,A),Ad(S_2,A))\ldots, Ad(S_k,A)),Ad(S_0,A'))$, where $S_i$, $i=0,1,\ldots, k$ are permutation matrices.

Edit: I now have the following conjecture regarding the Lie algebra: It is simply given by $\{A \in \mathfrak{u}(n): \sum_j A_{ij} = 0, \text{ for all }i \in [n]\}$, so it looks like the dimension should be $n^2 - 2n+1$. If so the group will be $U(n-1)$ acting on $V = \{z_1 + \ldots + z_n = 0\}$. The construction of explicit basis by Adjoint action remains.