= Solution
{c}
Conjugation by the <Hadamard gate> sends $(X,Y,Z)$ to $(Z,-Y,X)$. Therefore its <Bloch vector> action is
$$
\boxed{H:\ (s_x,s_y,s_z)\longmapsto(s_z,-s_y,s_x).}
$$
It is a rotation through $\pi$ about $(\mathbf e_x+\mathbf e_z)/\sqrt2$: $H=(X+Z)/\sqrt2$ differs by a <global phase> from the corresponding spin rotation.
The <S gate> and <T gate> differ by global phases from $z$-axis rotations:
$$
\boxed{S:\ R_z(\pi/2),\qquad T:\ R_z(\pi/4).}
$$
Indeed $\operatorname{diag}(1,e^{i\theta})=e^{i\theta/2}e^{-i\theta Z/2}$, which sends $(s_x,s_y,s_z)$ to $(s_x\cos\theta-s_y\sin\theta,s_x\sin\theta+s_y\cos\theta,s_z)$.
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