Suppose $K$ is a finite cyclic group, $H$ is an arbitrary group. Consider two homomorphisms $\phi_1, \phi_2: K \to \operatorname{Aut}(H)$ s.t $\phi_1(K), \phi_2(K)$ are conjugate in $\operatorname{Aut}(H)$.
Now let $\sigma \in \operatorname{Aut}(H)$ s.t $\sigma \phi_1(K) \sigma^{-1}=\phi_2(K)$.
Since $K=\langle x\rangle$ is cyclic, $\sigma \phi_1(x) \sigma^{-1}=\phi_2(x)^a$ for some $a\in \Bbb Z$ then for any $k \in K$ ; $\sigma \phi_1(k) \sigma^{-1}=\phi_2(k)^a$.
To prove that $\exists b\in \Bbb Z$ s.t $x^{ab}=x$.
I need this result to prove one problem from Dummit and Foote Sec $5.5$ Question no $6$. Please help, here in this post also I could not see the answer of my particular doubt.
