How many homomorphisms from $\mathbb Z_4$ to $S_4$? Let $\varphi: \mathbb{Z_4}\to S_4$ be a homomorphism.
From theory I have:

*

*Since $1_{\mathbb{Z_4}}$ is the identity element in $\mathbb{Z_4}$, $\varphi(1_\mathbb{Z_4})=1_{s_4}$.

*$\varphi(1^{-1})=\varphi(3)=(\varphi(1))^{-1}, \varphi(2^{-1})=\varphi(2)=(\varphi(2))^{-1}$
These don't seem enough to find all homomorphisms. What am I missing?
 A: There are problems with your calculation. You seem to be confusing the additive and multiplicative structure on $\mathbb{Z}_4$. If you say that the inverse of 1 is 3, then this means you consider the additive structure. But then the unit element is 0, not 1. 
But I can answer your question. Let $\mathbb{Z}_4$ be the ADDITIVE group of residue classes $\pmod 4$. Then the image of $1$ uniquely determines a homomorphism from $\mathbb{Z}_4$ to another group. This image can be any element whose order divides 4, and exactly those. So your question can be rephrased as follows: how many elements are there in $S_4$ whose order is 1, 2 or 4? 
There is exactly 1 element of order 1, the identity. 
There are $\binom{4}{2}= 6$ transpositions and 3 permutations that are products of two disjoint transpositions. So there are 9 elements of order 2. 
Finally, there are 6 elements of order 4 in $S_4$. 
So the answer is: there are $1+9+6= 16$ elements of order 1, 2 or 4 in $S_4$, hence 16 homomorphisms from $\mathbb{Z}_4$ into $S_4$. 
A: You can also think in terms of image of the homomorphism, according to the kernel (first homomorphism theorem).
For $\operatorname{ker}\phi=\{0\}$, you get the three copies of $\Bbb Z_4$ into $S_4$, namely the subgoups of $S_4$ generated by the $4$-cycles: $\langle(1234)\rangle$ (=$\langle(1432)\rangle$), $\langle(1243)\rangle$ (=$\langle(1342)\rangle$) and $\langle(1324)\rangle$ (=$\langle(1423)\rangle$). Each of them can be gotten in two ways, by swapping the two $4$-cycles in the subgroup as images of the generators $1$ and $3$. So, overall $3\times 2=6$ embeddings.
For $\operatorname{ker}\phi=\{0,2\}$, you get $\operatorname{im}\phi\cong C_2$: so, $0$ and $2$ are sent to $()$, while $1$ and $3$ are sent to one same element of order $2$ among the $9$ possible ones (namely: $(12)$, $(13)$, $(14)$, $(23)$, $(24)$, $(34)$, $(12)(34)$, $(13)(24)$, $(14)(23)$).
For $\operatorname{ker}\phi=\Bbb Z_4$, you get the trivial homomorphism: $i\mapsto ()$, for every $i\in \Bbb Z_4$.
Therefore, there are overall $6+9+1=16$ homomorphisms $\phi\colon \Bbb Z_4\to S_4$.
