= Solution
For general <spin> $S$, use <dimensionless> generators $S^j$ obeying $[S^i,S^j]=i\epsilon_{ijk}S^k$. The two exponentials represent the same group elements as for <spin> $1/2$, now in the $(2S+1)$-dimensional <representation> of $SU(2)$. A <representation> preserves products, so
$$
\boxed{e^{-i\Omega TS^z}e^{-i\Phi S^x}
=e^{-i\vartheta\hat{\mathbf n}\cdot\mathbf S}}
$$
with exactly the same axis and angle found in (c); after $N$ periods the angle is $N\vartheta$. One must not substitute higher-spin <matrices> into the two-term Pauli exponential in (b), which used a property special to <spin> $1/2$.
The dimension and <action> on <spin> states change, but the <spin-independent axis-angle composition> does not. In particular,
$$
\boxed{e^{-i2\pi\hat{\mathbf n}\cdot\mathbf S}=(-1)^{2S}I.}
$$
Integer-spin states return with positive sign and half-integer-spin states with negative sign. <spin> expectation <vectors> transform as spatial <vectors> in either case, while a general higher-spin state also carries higher multipole information.
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