The problem:
Given the three relations about square matrices $\alpha$ and $\beta$ $$\alpha^* n \beta^T - \alpha (n+1) \beta^{\dagger} = 1/2$$ $$\alpha^* n \alpha^T + \alpha (n+1) \alpha^{\dagger} = X$$ $$\beta^* n \beta^T + \beta (n+1) \beta^{\dagger} = P$$
where $n$ is a positive definite diagonal matrix and $X$ and $P$ are both positive definite real symmetric matrices, we need to show that $\alpha$ and $\beta$ can be expressed as $\alpha = \alpha_1 U$ and $\beta=\beta_1 U$, where U is a diagonal unitary matrix and $\alpha_1$ and $\beta_1$ are real matrices.
My progress:
I note that all the equations are invariant under transformations of the kind $\alpha \rightarrow \alpha V$, $\beta \rightarrow \beta V$, where V is a diagonal unitary matrix. Also, I can trivially see that $$\Im (\alpha (n+1/2)\beta^\dagger) = \Im(\alpha \alpha^\dagger) = \Im(\beta \beta^\dagger) = 0$$ The second equality strongly "suggests" that all rows of $\alpha$ have the same phase composition, i.e. phase of the $\{ij\}$ element only depends on $j$ $$\alpha_{ij} = |\alpha_{ij}| e^{i\xi_j}$$
The third equality "suggests" the same for $\beta$. If this hunch can can be shown to be true, then I see how the required statement is proved from here onwards. It is this hunch that I can't justify; I'm not a mathematician by trade. All help will be greatly appreciated. I encountered this problem in https://arxiv.org/abs/0905.2562 where the authors make this statement (right after eq (56)) without proof. An alternate proof which doesn't take the path I took is welcome too.