= Solution
A <cyclic group> has one <generator of a group>: $G=\langle a\rangle=\{a^j:j\in\mathbb Z\}$. An <abelian group> has $xy=yx$ for every pair of elements. Powers of one element commute because $a^ra^s=a^{r+s}=a^{s+r}=a^sa^r$, so \b[every <cyclic group> is <Abelian>]. The <Klein four-group> $C_2\times C_2$ is <Abelian> but is not a <cyclic group>: every nonidentity element has order two, whereas a <generator of a group> for a four-element <cyclic group> would have order four.
Fix a <generator of a group> $x$ of $C_n$. A <group homomorphism> is determined by $g=\phi(x)$ because $\phi(x^j)=g^j$, and the relation $x^n=1$ forces $g^n=1$. Conversely, such a $g$ defines $\phi(x^j)=g^j$: exponents differing by a multiple of $n$ give the same value, and addition of exponents verifies the <group homomorphism> law. Thus the <homomorphism from a finite cyclic group> correspondence is
$$
\boxed{\operatorname{Hom}(C_n,G)\ \longleftrightarrow\ \{g\in G:g^n=1\}.}
$$
For $S_4$, the order of a <permutation> is the <least common multiple> of its disjoint cycle lengths. The condition $\sigma^4=1$ allows identity, <transpositions>, two disjoint <transpositions>, and four-cycles. The \b[sixteen homomorphisms] are $\phi_\sigma(x^j)=\sigma^j$, with the full list of possible generator images
$$
\boxed{\begin{gathered}
1;\\
(12),(13),(14),(23),(24),(34);\\
(12)(34),(13)(24),(14)(23);\\
(1234),(1243),(1324),(1342),(1423),(1432).
\end{gathered}}
$$
The remaining eight elements of the <symmetric group> are <three-cycles> and do not qualify.
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