Consider the group $U(n)$ of unitary $n\times n$ matrices, as well as its subgroup $V$ of $n\times n$ matrices which are obtainable by permuting rows of diagonal matrices $\mathrm{diag}(a_1,\dots,a_n)$ with $|a_i|=1$.
It is clear that $V\subsetneq U(n)$. However, I am trying to come up with an argument why we cannot have $V\cong U(n)$ as groups. Note that the fact that $V\subsetneq U(n)$ does not imply this! For instance, we also have $2\mathbb Z\subsetneq\mathbb Z$, yet $2\mathbb Z\cong\mathbb Z$.
My idea was the following. For every $A\in V$, the matrix $A^{n!}$ is diagonal. Thus, if we have $A,B\in V$, then the two matrices $A^{n!}$ and $B^{n!}$ must commute. I expect this to fail for the matrices in $U(n)$. However, I am yet to find two matrices $A,B\in U(n)$ which do the job, meaning that $A^{n!}B^{n!}\neq B^{n!}A^{n!}$. Sure, if we fix some $n$, say $n=2$, then we can come up with such $A$ and $B$ explicitly. But how to solve the problem for general $n$?
Any ideas? I am also open for other strategies proving $V\not\cong U(n)$.
If $n\geq 3$, we can consider the following matrices. Let $$A_0=\begin{pmatrix} \cos a &-\sin a &\\ \sin a& \cos a&\\&& 1\end{pmatrix},\ B_0=\begin{pmatrix}1&&\\ &\cos b &-\sin b \\ &\sin b& \cos b\end{pmatrix},$$ where we assume $a,b$ are real, irrational multiples of $\pi$, so that for any integer $N$, $(\sin Na)(\sin Nb)\neq 0$. $A$ and $B$ will be the orthogonal sum of $A_0,B_0$ with the identity matrix $I_{n-3}$.
These $A,B$ are orthogonal, hence unitary as well. It is straightforward to see that $A^{n!}, B^{n!}$ are of the same form replacing the parameters $a,b$ with $n!a, n!b$ respectively. To see non-commuting property, look at the $(3,1)$-entry of the product matrices: $$(A^{n!}B^{n!})_{31}=0,\ (B^{n!}A^{n!})_{31}=\sin (n!a) \cdot \sin (n!b)\neq 0,$$ where the last $\neq$ follows from the preceding remark on the choice of $a,b$.