Let $G=K \ltimes N$ be a semi-product of two finite groups K and N with coprime order, N being the normal subgroup. My question is, if $K'$ is any subgroup of $G$ with $|K'|=|K|$, are $K'$ and $K$ conjugate? if not, does there exist $f\in \mathrm{Aut}(G)$ such that $f(K)=K'$?
Second Edit: Thanks to Derek Holt's comment below, the anwser to my first question and hence both the two are yes, by Schur-Zassenhaus Theorem.
Edit: As the comments below, maybe I should say something more on the motivation of this question. It arises in my mind when I am figuring out when two semidirect product $K\ltimes_{\phi_1} N$ and $K\ltimes_{\phi_2} N$ are isomorhic, where $\phi_i: K\to \mathrm{Aut}(N)$ are given. An easy sufficient condition is that $\phi_1 = \phi_2\circ \alpha$ or $\phi_1 = {\phi_2}^\beta$, or the combination of these two types; here $\alpha \in \mathrm{Aut}(K), \beta\in \mathrm{Aut}(N)$, and ${\phi_2}^\beta$ is the conjugate action of $\phi_2$.
When $K$ is a Sylow p-subgroup, then I can show that this is also a necessary condition for $K\ltimes_{\phi_1} N$ and $K\ltimes_{\phi_2} N$ being isomorphic. In general, I just wonder this holds or not when $|K|$ and $|N|$ are coprime, and it will hold if the anwser to my second question is yes. (Of course, if $K$ Sylow, then clearly the anwser is yes.) But, I can not make any real progress to my question.