Consider $G=P\ltimes Q$ where $P\cap Q=\{e\}$ and $Q<N_G(P)$. Here $\ltimes$ is the inner semidirect product.
Here I believe it is the case that the conjugation action of $P$ on $Q$ will determine the product structure of $G$.
So how do we know that the product structure on $G$ is determined by the conjugation action of $P$ on $Q$?
Is it correct to say that the automorphisms of $G$ determine the possible product structures, and so if we can construct a surjective map from the possible conjugation actions of $P$ on $Q$ to the automorphisms of $G$ then we will know that the conjugation action determines the product structure of $G$?
If so, what notation would express that map generally?
Can a concrete map be demonstrated for this case without referring to external semidirect products? I think that there's a generalization of this involving external semidirect products, but I'm trying to work up to that.
The context here is that I'm studying a classification of groups of order 6. I understand that classification pretty well in the concrete sense, but am trying to really get my head around the abstract ideas before moving on to harder examples.
Not sure if this is what you mean. For given $P,Q$, the semidirect group notation $P\ltimes Q$ is ambiguous unless you specify an action of $P$ on $Q$. (However, you wil often find the definition of the action omitted if it is "understood" or there is only one natural choice for a nontrivial action). Once you have that, $P, Q$ become subgroups of $P\ltimes Q$ (more formally: there exists a canonical embedding) and these subgroups have the property that $Q$ is normal and the original action of $P$ on $Q$ turns into conjugation. For the inner semidirect product we start at this end, i.e. the "secret" actoin of $P$ on $Q$ is already conjugation. Since $P,Q$ also generate $G$, multiplication in $G$ is indeed determined by this conjugation. A priori, "$P,Q$ generate $G$" merely means that each element of $G$ can be written as $p_1q_1p_2q\cdots p_nq_n$ with $p_i\in P, q_i\in Q$. If you know how to conjugate, you can shorten such products: If $n>1$, there is $p'$ with $q_{n-1}p_n=p'q_{n-1}$ and then replacing $p_{n-1}p'$ and $q_{n-1}q_n$ with single elements of $P$, $Q$ we obtain a shorter product. Hence ultimatley all elements of $G$ can be written in the form $pq$ and their multiplication $p_1q_1\cdot p_2q_2=p_3q_3$ is again governed by the permutation action.