Let G=Sym(n). Suppose that $\displaystyle \theta \in Hom(C, G)$, where C is a cyclic group. Then how to show that $\displaystyle \gamma (Ker \theta)=Ker \theta$ for $\displaystyle \gamma \in Aut(C).$
the claim is true for any group G. let $\displaystyle C=<x>$ and $\displaystyle \ker \theta = <x^m>.$ let $\displaystyle \gamma(x)=x^r$ and $\displaystyle x = \gamma(x^s).$ then for any integer $\displaystyle k$:
$\displaystyle \subseteq: \ \ \theta(\gamma(x^{km}))=(\theta(x^{km}))^r = 1_G.$
$\displaystyle \supseteq: \ \ x^{km}=\gamma(x^{skm}) \in \gamma(\ker \theta).$
forget about that proof! it can be done much easier and in general case:
let $\displaystyle N=<x^m>$ be any subgroup of $\displaystyle C=<x>$ and $\displaystyle \gamma \in \text{Aut}(C).$ let $\displaystyle \gamma(x)=x^r.$ then $\displaystyle \gamma(x^{km})=x^{rkm} \in N.$ thus $\displaystyle \gamma(N) \subseteq N.$ replacing $\displaystyle \gamma$ with $\displaystyle \gamma^{-1}$ gives us $\displaystyle N \subseteq \gamma(N).$ thus $\displaystyle \gamma(N)=N.$