This is problem 2.6.P16 from Matrix Analysis by Horn and Johnson. The problem is: given $U,V\in M_n$ unitary, then there exist $X,Y,D\in M_n$ unitary, with $D$ additionally diagonal, such that $U = XDY, V = Y^*DX^*$.
My attempt uses the following lemma, proven in an earlier problem:
Lemma 1 Let $A,B\in M_{n,m}$, then $AB^*$ and $B^*A$ are both normal if and only if there exist unitary $X\in M_n, Y\in M_m$, such that $A = X\Sigma Y^*$ and $B = X\Lambda Y^*$ for some diagonal $\Sigma, \Lambda \in M_{n,m}$, and $\Sigma$ has the singular values of $A$ on its main diagonal.
Now, $UV$ and $VU$ are both unitary, and hence normal, so by lemma 1, there exist $X,Y\in M_n$ unitary, such that $$ U = X\Sigma Y,\qquad V^*=X\Lambda Y\implies V = Y^*\overline{\Lambda} X^*, $$ however I fail to see how to conclude that $\Sigma=\Lambda=D$. However, I do notice that $\Sigma = I_n$, since $U$ is unitary, and the singular values of a normal matrix are the absolute values of the eigenvalues, which have magnitude $1$, so in fact $U = XY$, but I fail to see how that helps me.
EDIT: I think there is definitely an error in the book. Say $U = P\Sigma Q^* = PQ^*$ for some unitary $P,Q$, then if we take $\hat{V}^*= QV^*P^*$, we have that $\hat{V}^*$ is unitary with spectral decomposition $\hat{V}^* = Z^*\Lambda Z$. Take $X = PZ, Y=Z^*Q^*$, then $X\Sigma Y=XY=PQ^*=U$, and $Y^*\overline{\Lambda} X^*=QZ\overline{\Lambda}Z^*P^*=Q\hat{V}P^*=V.$ But then $\Lambda$ need not be real, thus need not be $I_n$.
On the other hand, if we take the SVD of $\hat{V}^* = R^*\Sigma_V T = R^*T$ instead of the spectral decomposition, where $R,T$ are unitary, and set $X = PR^*, Y = TQ^*$, then $$ Y^*\Sigma_V X^*= Y^*X^* = QT^*RP^*=Q\hat{V}P^*=V, $$ but $$ X\Sigma Y= XY = PR^*TQ^*= PV^*Q^*\neq U. $$
So I guess my main question is: what is the correct statement, if any.