Find all homomorphisms $\varphi : \Bbb Z_6 \to S_3$. [duplicate]

Let $f:C_8\rightarrow S_3$ be non-trivial homomorphism. Then one can see that $\frac{C_8}{Kerf}$ is isomorphic to a subgroup of $S_3$. Note that $|\frac{C_8}{Kerf}|$ is power of $2$. Thus, $\frac{C_8}{Kerf}\cong \langle \sigma \rangle$ where $o(\sigma)=2$. Now, Assume that $C_8=\langle x \rangle$. We can define homomorphisms $f_1(x)=(1,2)$, $f_2(x)=(1,3)$ and $f_3(x)=(2,3)$. However, homomorphisms from $C_6$ to $S_3$ are little different. Suppose that $f:C_6\rightarrow S_3$ is homomorphism. By similar argument, $\frac{C_6}{kerf}$ is isomorphic to a subgroup of $S_3$. As you know, since $S_3$ is a non-abelian group and $C_6$ is an abelian group, $ker f\neq 1$. Hence, $|\frac{C_6}{kerf}|=2$ or $3$.

Case (I): Let $|\frac{C_6}{kerf}|=2$ and $C_6=\langle x\rangle$. Then one can see that $f_1(x)=(1,2)$, $f_2=(2,3)$ and $f_3=(1,3)$ are homomorphisms.

Case (II): Let $|\frac{C_6}{kerf}|=3$ and $C_6=\langle x\rangle$. Then one can see that $f_4(x)=(1,2,3)$, $f_5=(1,3,2)$ are homomorphisms.

We can conclude that $|Home(C_6,S_3)|=6$ and $|Home(C_8,S_3)|=4$.