Let $n$ be a natural number and let $A = (a^{i}_{j})_{1 \leq i,j \leq n} \in U_{n}$, i.e., $A^{*}A = AA^{*} = I_{n}$. Suppose for each choice of indices $i,j,k,\ell \in \{1,\dots,n\}$ it holds $a^{i}_{j}(a^{k}_{\ell})^{*} = (a^{i}_{j})^{*}a^{k}_{\ell}$, where upper indices are row indices, lower indices are column indices and "$^{*}$" denotes complex conjugation.
From elementary considerations it follows that for all choices of indices $i,j,k,\ell \in \{1,\dots,n \}$ it holds $\mathrm{arg}(a^{i}_{j}) = \mathrm{arg}(a^{k}_{\ell})$ or $\mathrm{arg}(a^{i}_{j}) = \mathrm{arg}(a^{k}_{\ell}) + \pi$.
I assume that in the situation described, there are a matrix $B \in O_{n}$ and a complex number $\zeta$ such that $A = \zeta B$. I do not know how to show this, however. Suggestions?
Since $AA^\ast=I$, some $a_{ij}$ is nonzero. Let $\omega=\frac{\overline{a}_{ij}}{|a_{ij}|}$. By assumption, $\omega a_{kl}=\frac{\overline{a}_{ij}a_{kl}}{|a_{ij}|}$ is real. Hence $Q=\omega A$ is a real matrix and $QQ^T=QQ^\ast=(\omega A)(\omega A)^\ast=AA^\ast=I$, i.e. $A=\overline{\omega}Q$ where $|\overline{\omega}|=1$ and $Q$ is real orthogonal.