 Solution

The two density operators have the same eigenvalues $a,a,b,b$, so a permutation of basis vectors makes them unitarily equivalent. A dynamical equivalence must additionally respect the symplectic dynamical symmetry in quantum control.

Define the symplectic dual $\tilde{\rho }=J{\rho }^T{J}^{†}$. For a unitary $U$ preserving $J$, both $U^{T}JU=J$ and $UJ{U}^T=J$ hold. In particular $J{U}^{\ast }=UJ$, and thereforeIt follows by cyclicity of the trace that the symplectic invariant of a density operatoris unchanged by every reachable unitary conjugation. The given $J$ reverses the diagonal order, soConsequentlyHence the states are not dynamically equivalent, despite their identical ordinary spectra. This works for all the allowed values, including $a=0$, and for any subgroup satisfying the symmetry in part (c).