= 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 $\widetilde\rho=J\rho^TJ^\dagger$. For a unitary $U$ preserving $J$, both $U^TJU=J$ and $UJU^T=J$ hold. In particular $JU^*=UJ$, and therefore
$$
\widetilde{U\rho U^\dagger}=J U^*\rho^T U^T J^\dagger=U\widetilde\rho U^\dagger.
$$
It follows by cyclicity of the <trace> that the <symplectic invariant of a density operator>
$$
I_J(\rho)=\operatorname{Tr}(\rho\widetilde\rho)
$$
is unchanged by every reachable <unitary conjugation>. The given $J$ reverses the diagonal order, so
$$
\widetilde\rho_0=\operatorname{diag}(b,b,a,a),\qquad\widetilde\rho_1=\operatorname{diag}(a,b,b,a)=\rho_1.
$$
Consequently
$$
I_J(\rho_0)=4ab,\qquad I_J(\rho_1)=2(a^2+b^2),\qquad I_J(\rho_1)-I_J(\rho_0)=2(a-b)^2>0.
$$
Hence \b[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).
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