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$$PXP^{-1}=Y$$ $$PYP^{-1}=X$$ $$QXQ^{-1}=X$$ $$QYQ^{-1}=-Y$$ from which it follows that $$P(X+iY)P^{-1}=Y+iX=i(X-iY)$$ $$Q(X+iY)Q^{-1}=X-iY$$ This shows that $X+iY$ is conjugate to $i(X+iY)$, so its eigenvalues …

I will show that it is not possible for $\phi=\pi/2$, so it is certainly not for general $\phi$. (actually, I don't think that it is possible for any single $\phi$ except $0$ and $\pi$, by an analogou …

Thus the eigenvalues of $A$ are those of $\prod N_{\mu_i}$ plus those of $\overline{\prod N_{\mu_i}}$, which are their complex conjugates. … Moreover, since $N_\mu$ has determinant $1$, so does $\prod N_{\mu_i}$, so its two eigenvalues are inverses of each other. …

Some computation in sage yielded the following example, with $n=8$ and $P$ the cyclic permutation $(12345678)$: $$M=\left( \begin{array}{cc} 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 & …