Relationship between internal and external semidirect products. $\newcommand{\Aut}{\text{Aut}}$
I'm trying to distinguish the two in my mind although they seem to be the same. Here are the definitions I'm working with.

Definition:    Let $G$ be a group with subgroups $H,N \leq G$ and $N \unlhd G$. We say that $G$ is an internal semidirect product of $H$ and $N$, denoted $N \rtimes H$ or $H \ltimes N$, if $G = HN = NH$ and $H \cap N = \{1\}$.
Definition:    Let $H$ and $N$ be groups and $\varphi:H \rightarrow \Aut(N)$ be a homomorphism. The external semidirect product of $N$ and $H$ w.r.t $\varphi$; denoted by $N \rtimes_\varphi H$, is the set $N \times H$ with binary operation
$$
 (n_1,h_1)(n_2,h_2) : = (n_1 \varphi(h_1)(n_2),h_1h_2).
 $$

So I understand that when you look at the definition of an internal semidirect product, it's almost "implying" (might not be the best term, informing maybe?) the definition of the external direct product. This is because since $N \unlhd G$ then for any $h \in H$ conjugation of $N$ by $h$ is an automorphism of $N$ and so $\varphi_h = hnh^{-1}$ constitutes an automorphism of $N$. Moreover for any $h$ the map $\psi:H \rightarrow \Aut(N)$ by $h \mapsto \varphi_h$ is a homomorphism called the conjugation homomorphism of $N \rtimes H$. Then using this we can "hide" the group operation in $G$ by the group operations in $N$, $H$ and the conjugation automorphism by
\begin{align*}
  (n_1h_1)(n_2h_2) &= n_1h_1n_2h_2 \\
  & = n_1h_1 n_2h_1^{-1}h_1h_2 \\
  & = n_1(\varphi_{h_1}(n_2))h_1h_2 \\
  & = n_1 \psi(h_1)(n_2)h_1h_2. 
 \end{align*}
So fair enough this basically has given us what the binary operation on $N \rtimes_\varphi H$ should be. But then if we set $N' = N \times 1_H$ and $H' = 1_N \times H$ we could show that $N \rtimes_\varphi H = N' \rtimes H'$. So the external semidirect product of $N$ and $H$ is just the internal semidirect product of $N'$ and $H'$ which are isomorphic to $N$ and $H$ respectively.
So I understand that at a basic level these are different since one contains tuples and the other points, and maybe the distinction is that an internal semidirect product is something you check that a group $G$ might be in hiding, but an external semidirect product is a way to build up new groups.
Is this a valid takeaway for this construction? In particular what makes them worth distinguishing if the external semidirect product is an intenral semidirect product of groups isomorphic to $N$ and $H$? Thanks in advance for your help.
 A: With the external construction you can create new groups you may not have seen before, such as building the nonabelian group of order $12$ of the form $\mathbf Z/(3) \rtimes_{\varphi} \mathbf Z/(4)$ where $\mathbf Z/(4)$ acts on $\mathbf Z/(3)$ by negation: $(a,b)(c,d)= (a + (-1)^bc,b+d)$. That is not $A_4$ or $D_6$, which are the familiar nonabelian groups of order $12$: it has a cyclic $2$-Sylow subgroup, while $A_4$ and $D_6$ have noncyclic $2$-Sylow subgroups.
With the internal construction you can recognize groups you know as having this kind of structure, which is nothing other than having a normal subgroup $N$ that admit a complementary subgroup $H$ ($G = NH$ and $N \cap H = \{1\}$.) For instance, the dihedral group $D_n$ of order $2n$ is abstractly of the form $\mathbf Z/(n) \rtimes \mathbf Z/(2)$, where $\mathbf Z/(2)$ acts on $\mathbf Z/(n)$ by inversion ($(a,b)(c,d) = (a+(-1)^bc,b+d)$).
If you want to classify all groups of a particular order, both the internal and external constructions can be useful: in some cases you can show all the groups of a particular order occur as a semidirect product of nontrivial groups, and then to give concrete models of the different types of groups up to isomorphism you may want to show that concrete groups have the desired type of semidirect product structure.  See Sections $4$ and $7$ here. (In that file, the notation for semidirect products is $H \rtimes K$, so $H$ denotes the normal subgroup and $K$ is a complementary subgroup, which does not match the way you use the notation $H$ above.)
