Given $A \in \mathbb{R}^{n \times n}$, with $0 < |\det(A)| < 1$. Does a diagonal matrix $D$ exist such that
$$ B = D^{-1} A D $$ is the principal submatrix of an $(n+1)\times(n+1)$ orthogonal matrix, i.e., $$ X = \begin{bmatrix} B & b \\ c & d \end{bmatrix} \textrm{ with } X^T X = I. $$
This is equivalent to $B$ having singular values all 1 except one, i.e., $$ \sigma(B) = [1, 1, \dots, 1, |\det(A)| ] = s^*. $$ There is, of course, a non-diagonal $D$ which satisfies the problem for any $A$. For diagonal $D$, however, not all $A$ do have a solution, e.g., a diagonal $A$ with $\sigma(A) \neq s^*$. I'd be interested also in a numerical algorithm to find $D$ (if it exists).