Let $S(n)$ be the (unitary) matrix group of $n\times n$ permutation matrices. This is clearly a finite group of order $n!$.

Let $U_n\in U(n)$ be a unitary matrix which is not a (unitary) monomial matrix (sometimes called generalized symmetric matrices). In other words, $U_n$ is not a product of a permutation and a diagonal matrix.

If the addition of any non-monomial matrix $U_n$ as a generator with $S(n)$ ($\langle S(n), U_n \rangle$) always generates an infinite order group I refer to the group $S(n)$ as maximal.

Can we prove the existence of maximal groups $S(n)$ for some $n$?

Can we prove that there exists an integer $k$ such that for all $n>k$ and any non-monomial matrix $U_n$, that $\langle S(n), U_n \rangle$ necessarily generates an infinite order group?

I can show that $k>4$ see the $H\otimes H$ example in the comments to this question: Are generalized symmetric groups maximal finite groups (in a certain sense)?