= Solution
{c}
A <group homomorphism> from an additive <cyclic group> $\mathbb Z_n$ is determined by the image $a$ of $1$, subject exactly to $a^n=e$ in multiplicative notation, or $na=0$ in an additive target.
For $\mathbb Z_{18}\to\mathbb Z_{15}$, the relation is $18a\equiv0\pmod{15}$, equivalently $3a\equiv0\pmod{15}$. The possibilities are $a=0,5,10$, and each defines a <homomorphism between cyclic groups>. Thus there are \b[three] such maps, an instance of the <number of homomorphisms between finite cyclic groups>.
For $\mathbb Z_9\to S_4$, the image of $1$ must be a <permutation> whose <order of a group element> divides $9$. A <permutation> on four symbols can have order $1,2,3$ or $4$, so the allowed images are the identity and the <three-cycles>. There are $\binom43(3-1)!=8$ <three-cycles>: choose their three symbols and one of the two cyclic orientations. Hence there are \b[nine] maps.
Finally, each <transposition> in $S_5$ has order $2$, so its image in $\mathbb Z_5$ must have order dividing both $2$ and $5$, and is therefore the identity. Every <permutation> is a product of <transpositions>, so every element must map to the identity. There is \b[one] map. In the order asked, the counts are
$$
\boxed{3,\qquad9,\qquad1.}
$$
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