Suppose $U$ is a unitary matrix. Does there exist a non-singular matrix $R$ such that $$ U=\bar{R}R^{-1}? $$ The bar denotes the complex conjugation of each entry of the matrix $R$.
Remarks: 1) It is obviously true for $1\times 1$. 2) If $U$ admits the decomposition $U=VDV^{T}$, where $V$ is orthonormal and $D$ diagonal with diagonal entries of unit modulus, then $R:=VD^{-1/2}V^{T}$.
The required $R$ exists if and only if $U$ is complex symmetric.
If $U=\overline{R}R^{-1}$, then $UR=\overline{R}$. Hence $U\overline{UR}=U\overline{\overline{R}}=UR=\overline{R}$. In turn, $U\overline{U}=I$ and $\overline{U}=U^\ast$. Hence $U=U^T$.
Conversely, if $U$ is complex symmetric, it admits a special form of singular value decomposition $V\Sigma V^T$, known as Takagi factorisation. Since $U$ is unitary, its singular values are all equal to $1$. Hence $\Sigma=I$ and $U=VV^T$. Now take $R=\overline{V}$ and we are done.
