For $C_8$ to $S_3$: Let $C_8$ be generated by $x$. Then $C_8 = \langle x : x^8 = e \rangle = \{e,x,x^2,x^3,x^4,x^5,x^6,x^7\}$. But how do I show homormorphisms? I'm supposed to pick elements of $x$ such that its eighth powers will be $e$, but how? Is this process similar for $C_6$?
I do know, however, that there are no isomorphisms, as isomorphisms preserve structure and $S_3$ is not a cyclic group unlike $C_8$. The two groups are not even of the same order. As for $C_6$ to $S_3$, no isomorphism exists as well, because even though they have same order, one of them is cyclic and the other is symmetric.
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 $|Home(C_8,S_3)|=4$.