Step 1: List every possible element order in $S_5$ directly.
An element of $S_5$ is determined by its cycle type: a single 5-cycle, a 4-cycle plus a fixed point, a 3-cycle plus a 2-cycle, a 3-cycle plus two fixed points, two 2-cycles plus a fixed point, a single 2-cycle plus three fixed points, or the identity. Taking the lcm of cycle lengths in each case gives orders $5, 4, 6, 3, 2, 2, 1$. No value exceeds $6$, and $6$ does occur (from $(123)(45)$), so the maximum order in $S_5$ is exactly $6$. Statement (I) holds.
Step 2: Use the abelianization of $S_3$ instead of listing normal subgroups.
Any homomorphism from a group $G$ to an abelian group factors through its abelianization $G/[G,G]$. For $S_3$, the commutator subgroup is $A_3=\{e,(123),(132)\}$, so\[S_3/[S_3,S_3] = S_3/A_3 \cong \mathbb{Z}_2\]Hence any homomorphism $f:S_3\to \mathbb{Z}_6$ factors as $S_3 \twoheadrightarrow \mathbb{Z}_2 \xrightarrow{\ \bar f\ } \mathbb{Z}_6$, and its image is the image of $\bar f$.
Step 3: Determine the possible images.
A homomorphism out of $\mathbb{Z}_2$ sends the generator to an element of order dividing $2$ in $\mathbb{Z}_6$, namely $0$ or $3$. So the image is either trivial (order $1$) or $\{0,3\}$ (order $2$). No choice gives order $3$, so statement (II) fails.
Step 4: Conclusion.
(I) is true and (II) is false.\[\boxed{\text{Only (I)}}\]