Unitary group is conjugacy-closed in general linear group

Statement
Let $$GL(n,\mathbb{C})$$ denote the general linear group: the group of invertible $$n \times n$$ complex matrices. Let $$U(n,\mathbb{C})$$ denote the unitary group: the subgroup comprising matrices $$A$$ such that $$AA^*$$ is the identity matrix. Then, $$U(n,\mathbb{C})$$ is conjugacy-closed in $$GL(n,\mathbb{C})$$: any two unitary matrices that are conjugate over $$GL(n,\mathbb{C})$$, are conjugate in $$U(n,\mathbb{C})$$.

Proof
The proof uses the following facts:


 * By the spectral theorem for unitary matrices, any unitary matrix is conjugate, in the unitary group, to a diagonal unitary matrix.
 * Also, clearly any two diagonal unitary matrices that are conjugate in the general linear group, are conjugate by a permutation matrix, hence they are conjugate in the unitary group.

Thus, we have established a conjugate-dense subgroup of $$U(n,\mathbb{C})$$ (namely the diagonal unitary matrices) such that any two elements of the subgroup are conjugate in $$GL(n,\mathbb{C})$$ iff they are conjugate in $$U(n,\mathbb{C})$$. This shows that $$U(n,\mathbb{C})$$ is conjugacy-closed in GL(n,\mathbb{C}).