Semi-Direct Product Unique up to a Isomorphism My professor mentioned this theorem (which he said is in Dummit and Foote but I couldn't find). 

Given groups $H$ and $K$ and a homorphism $\varphi : H \rightarrow \text{Aut}(K)$, there is up to an isomorphism one semidirect product $G=K \rtimes _\varphi H$ such that the conjugation action of $H$ on $K$ is given by $\varphi$.

I don't understand the point of this statement, is it saying for every possible conjugation action of $H$ and $K$ we can only construct one group, so we can construct as many groups as there are inner automorphism of $K$??
 A: The theorem could perhaps be better stated as follows:
Theorem Let $H$ and $K$ be groups and let $\varphi : H \to \operatorname{Aut}(K)$ be a homomorphism. There is a unique group $G$, up to isomorphism, such that $K$ and $H$ are subgroups of $G$, $G$ is the internal semidirect product of $K$ and $H$, and the conjugation action of $H$ on $K$ is $\varphi$.
This is easy to prove, so you should definitely give it a shot!
Notably, this does not mean that each map $\varphi$ gives a different isomorphism class of group $K \rtimes_\varphi H$! For example, there is only one group of order $21$ (up to isomorphism), but there are $6$ distinct actions $\mathbb{Z}/3\mathbb{Z} \to \operatorname{Aut}(\mathbb{Z}/7\mathbb{Z})$! Thus, each of these six actions constructs an isomorphic group of the form $\mathbb{Z}/7\mathbb{Z} \rtimes \mathbb{Z}/3\mathbb{Z}$.
In other words, if two semidirect products of $K$ and $H$ have the same conjugation action, then they must be isomorphic, but isomorphic semidirect products need not have the same conjugation action.
