= Solution
For a Hermitian <Quantum Hamiltonian>, $[\sigma_m,\sigma_n]^\dagger=-[\sigma_m,\sigma_n]$. Consequently
$$
L_{mn}^*=-i\operatorname{Tr}([\sigma_m,\sigma_n]^\dagger H)
=i\operatorname{Tr}(H[\sigma_m,\sigma_n])=L_{mn}.
$$
Swapping $m,n$ changes the sign of the <commutator>, so $L_{nm}=-L_{mn}$. The generator is therefore real and antisymmetric. Its last column vanishes because the identity commutes with $H$, so $c=0$ and the traceless block satisfies $A^T=-A$.
For constant $H$, the <Bloch vector> evolves by $s(t)=R(t)s(0)$ with $R(t)=e^{At}$. Direct differentiation gives
$$
\frac d{dt}(R^TR)=R^T(A^T+A)R=0,
\qquad R(0)=I.
$$
Hence $R^TR=I$. Its <determinant> is continuously connected to one and cannot change between the two possible values $\pm1$, so $\det R=1$. This proves that the evolution is a rotation in the <special orthogonal group>. A time-dependent Hermitian <Quantum Hamiltonian> has the same conclusion from $\dot R=A(t)R$.
These <Hamiltonian rotations of Bloch vectors> preserve the Euclidean norm and <purity of a density operator>. For $N>2$, not every rotation of the ambient sphere is a physical unitary conjugation: the higher-dimensional positivity constraints and full <spectrum> of the <density operator> impose further restrictions.
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