= Solution
Write the <unitary conjugation> as $\rho(t)=U(t)\rho(0)U(t)^\dagger$. The identity component is unchanged, and the generalized <Bloch vector> transforms by
$$
s_k(t)=\sum_\ell O_{k\ell}(U(t))s_\ell(0),\qquad O_{k\ell}(U)=\operatorname{Tr}(\sigma_kU\sigma_\ell U^\dagger).
$$
These coefficients are real, by the same trace argument as in part (a). <Unitary conjugation> preserves the <Hilbert-Schmidt inner product>, so the conjugated traceless generators form another <orthonormal basis>. It follows that $O(U)^TO(U)=I$. The <unitary group> is connected, $O(I)=I$, and the <determinant> of an orthogonal <matrix> is $\pm1$; hence $\det O(U)=1$. Thus \b[Hamiltonian evolution rotates the generalized Bloch vector by an element of $SO(N^2-1)$].
Equivalently, in units $\hbar=1$, the <Von Neumann equation> gives
$$
\dot s_k=\sum_\ell A_{k\ell}s_\ell,\qquad A_{k\ell}=-i\operatorname{Tr}(\sigma_k[H,\sigma_\ell]).
$$
Cyclicity of the <trace> shows $A_{k\ell}=-A_{\ell k}$, so $A$ is a real <skew-symmetric matrix>. This is the generator form of <Hamiltonian rotations of Bloch vectors>, valid also for a time-dependent <Quantum Hamiltonian>. For $N>2$, Hamiltonians generally realize only a proper subgroup of all rotations of this real space; they preserve the whole density-operator <spectrum>, not merely its <purity of a density operator>.
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