= Solution
The <order of a group element> $g$ is the least <positive integer> $n$ such that $g^n=e$, where $e$ is the <identity element>; its order is infinite if no such $n$ exists.
Let $g$ have finite order $n$. A <group homomorphism> preserves the <group operation> and the identity, so
$$
\phi(g)^n=\phi(g^n)=\phi(e)=e.
$$
The order of an element divides every positive exponent that gives the identity. Hence
$$
\boxed{\operatorname{ord}(\phi(g))\mid\operatorname{ord}(g)}.
$$
If $\phi$ is a <surjective function> and $h\in H$ has order $m$, choose $g\in G$ with $\phi(g)=h$. The first result gives $m\mid n$, where $n=\operatorname{ord}(g)$. The element
$$
g^{n/m}
$$
then has order $m$, because the order of $g^k$ is $n/\gcd(n,k)$.
A <group homomorphism> $C_9\to S_4$ is determined by the image of a <generator of a group>[generator] of the <cyclic group> $C_9$, and that image must have order dividing $9$. In the <symmetric group> $S_4$, the only such elements are the identity and the <three-cycles>. There are
$$
\binom43(3-1)!=4\cdot2=8
$$
three-cycles. Therefore the number of homomorphisms is
$$
\boxed{1+8=9}.
$$
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