Let \(\displaystyle f,g :\mathbb{Z}/5\mathbb{Z} \to S_{5}\) be 2 non trivial group homomorphisms. Prove that there exists a \(\displaystyle \sigma \in S_{5}\) such that \(\displaystyle f(x)=\sigma g(x) \sigma^{-1}\) for every \(\displaystyle x \in \mathbb{Z}/5\mathbb{Z}\)

As tonio hinted, f and g are embeddings, i.e., the images of f and g are cyclic subgroups of order 5 in S_5.

Let A be the image of f such that A=<(a_1, a_2, a_3, a_4, a_5)> and let B be the image of g such that B=<(b_1, b_2, b_3, b_4, b_5)>. Since the generator of A and B have the same cycle type, they are in the same orbit under the action of conjugation. That means, there exists \(\displaystyle \sigma \in S_5\) such that \(\displaystyle \sigma b \sigma^{-1}=a\), where b and a are generators of B and A, respectively.

Once we find a \(\displaystyle \sigma\), the bijection is established between B and A. Think of a \(\displaystyle \sigma\) relabels the generator of B to the generator A. Then each element of B corresponds to A in a well-defined manner.