Relation Between Two Homomorphisms Let $f,g: \mathbb Z_5\to S_5$ be two non-trivial homomorphisms.  

Prove that for every $x\in\mathbb Z_5\,$ there exists $\,\sigma \in S_5\,$ such that $\,f(x)=\sigma g(x)\sigma^{-1}.$

I have found that $f,g$ are injective, but cannot proceed any further. 
Thanks for any help.
 A: Consider (1) where the generator can be sent to, (2) what their cycle type must look like, and (3) show that all possible images for the generators are conjugate. Now, (4) why is this sufficient?

Step One:

 Let $x$ be the generator for the cyclic group $C_5$ (written multiplicatively). So $x^5=e$. Observe that $f(x)^5=f(x^5)=f(e)=e$ (two different $e$s here, technically). And similarly for $g(x)$. Thus the image of $x$ under $f,g$ must satisfy $\sigma^5=e$. Furthermore we cannot have $\sigma=e$ because that would make the homomorphism trivial, as $f(x^n)=f(x)^n=e^n=e$ for all $x^n\in C_5$. Therefore we may conclude that $f(x)$ and $g(x)$ must be elements of order $5$.

Step Two:

 If $\sigma$ has order $5$ in $S_5$, the cycles in its disjoint cycle decomposition must have lcm = $5$, which means each such cycle has order $5$, and there cannot be more than one such cycle because that would imply that there are $\ge10$ numbers being acted on by $S_5:={\rm Perm}(\{1,2,3,4,5\})$. Thus we conclude that $f(x)$ and $g(x)$ are $5$-cycles.

Step Three:

 Know the formula $\sigma(a_1~a_2~\cdots~a_m)\sigma^{-1}=(\sigma(a_1)~\sigma(a_2)~\cdots~\sigma(a_m))$: thus, conjugating a cycle in a symmetric group by $\sigma$ amounts to the same thing as applying $\sigma$ to the numbers that appear in its cycle representation. Use this to conclude (in particular) any two $5$-cycles are conjugate.

Step Four:

 Conclude $f(x^n)=f(x)^n=(\sigma g(x)\sigma^{-1})^n=\sigma g(x)^n\sigma^{-1}=\sigma g(x^n)\sigma^{-1}$ for all $x^n\in C_5$ for some $\sigma\in S_5$, thus any two nontrivial homomorphisms are conjugate.

