 Solution

The spin commutation relations give $[S_{\mathrm{tot}}^z,S_n^{\pm }]=\pm S_n^{\pm }$. Using the product rule for a commutator,and the conjugate term also vanishes. The longitudinal field commutes with every $S_n^z$, so $[H,S_{\mathrm{tot}}^z]=0$. The Hamiltonian is invariant under the global spin rotation $U(\theta )=e^{-i\theta S_{\mathrm{tot}}^z}$: its continuous symmetry group is $U(1)$. Simultaneous energy and total-magnetization eigenstates can be chosen, and evolution preserves each fixed-magnetization sector. Local magnetization can still flow between sites; generally $[H,S_n^z]\ne 0$. In the fermion representation the conserved number will be $N_f=N/2-S_{\mathrm{tot}}^z$.