= Solution
{c}
By the preceding relabelling formula, every conjugate of $(123)$ is a <three-cycle>, and any <three-cycle> on four symbols can be obtained by a suitable relabelling. Therefore
$$
\boxed{C((123))=\{(123),(132),(124),(142),(134),(143),(234),(243)\}.}
$$
There are eight elements, agreeing with the count $\binom43\cdot2$.
Likewise, conjugation preserves the pattern of two disjoint <transpositions>, and every partition of four symbols into two pairs is realized by relabelling. Thus
$$
\boxed{C((12)(34))=\{(12)(34),(13)(24),(14)(23)\}.}
$$
These have the respective <cycle types> $(3,1)$ and $(2,2)$ in the <symmetric group> $S_4$.
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