Let G=Sym(n) and $\displaystyle \theta \in Hom(C, G)$ where C is a cyclic group. Since C is cyclic $\displaystyle \gamma (Ker \theta)=Ker \theta$, where $\displaystyle \gamma \in Aut(C)$ and the generators of $\displaystyle \theta(C)$ in G have the same cycle type. Therefore the map $\displaystyle \tilde \theta :N_{G}(\theta (C))\longrightarrow Aut(C/Ker \theta)$ is an isomorphism.

Can u explain the last sentence?

And Is this true for all groups? or is there a special group?