Is there a way to define other homorphisms, different from the conjugation mapping, in the definition of outer semidirect-products? I've found the following definition of the outer semidirect product on wikipedia

Let us now consider the outer semidirect product. Given any two groups
$N$ and $H$ and a group homomorphism $φ: H \to \operatorname{Aut}(N)$,
we can construct a new group $N \rtimes_\varphi H$, called the outer
semidirect product of $N$ and $H$ with respect to $\varphi$, defined
as follows:

*

*The underlying set is the Cartesian product $N \times H$

*The group operation $\bullet$ is determined by the homomorphism $\varphi$:
\begin{align}   
\bullet : (N \rtimes_\varphi H) \times (N
\rtimes_\varphi H) &\to N \rtimes_\varphi H\\
                                  (n_1, h_1) \bullet (n_2, h_2) &= (n_1 \varphi(h_1)(n_2),\, h_1 h_2) = (n_1 \varphi_{h_1}(n_2),\, h_1
 h_2) \end{align} for $n_1, n_2 \in N$ and $h_1, h_2 \in H$.


Question
I know from the definition of the inner semidirect product, that $\varphi$ maps some $h$ to  the conjugation of $N$ with $h$. But in the definition of the outer semidirect product, it seems like $\varphi$ could be any group homomorphism.
If I am right, assuming $\varphi : H \to \operatorname{Aut}(N)$ can be any group homomorphism, and it especially may be something different than the mapping to conjugations, I'd be interested in the definition of such a $\varphi$. Are there any examples?
 A: The answer to the question in the subject line in fact is no: a  semidirect product turns arbitrary automorphisms of $N$ into conjugations on $N$ from within a larger group. The external semidirect product is not giving you something for nothing: the automorphisms on $N$ from $H$ within $H \rtimes_\varphi N$ are conjugations, just like the internal semidirect product.
For an internal semidirect product $G = NH$, where $N \lhd G$ and $N \cap H = \{1\}$, the action of $H$ on $N$ is conjugation by elements of $H$ on $N$. Elements of $H$ need not commute with elements of $N$, so the conjugation of $H$ on $N$ can be nontrivial even if $N$ is abelian (making conjugation of $N$ on itself trivial).
For an external semidirect product $N \rtimes_\varphi H$, where $\varphi \colon H \to {\rm Aut}(N)$ is an arbitrary homomorphism and
$$
(n_1,h_1)(n_2,h_2) = (n_1\varphi_{h_1}(n_2),h_1h_2), 
$$
we view $N$ and $H$ inside $N \rtimes_\varphi H$ as the subgroups $\{(n,1) : n \in N\}$ and $\{(1,h) : h \in H\}$. For each $h \in H$, the automorphism $\varphi_h$ on $N$ is a conjugation by $(1,h)$ on $N$ within $N \rtimes_\varphi H$: for $n \in N$,
$$
(1,h)(n,1)(1,h)^{-1} = (\varphi_h(n),h)(1,h^{-1}) = (\varphi_h(n)\varphi_h(1),hh^{-1}) = (\varphi_h(n),1).
$$
So for an arbitrary homomorphism $\varphi \colon H \to {\rm Aut}(N)$, the action of $H$ on $N$ inside the external semidirect product $N \rtimes_\varphi H$ is a conjugation $H$ on $N$, just like for internal semidirect products.
Example.  Let $A$ be abelian, so it has the automorphism $a \mapsto -a$ of order $2$ (well, this has order $2$ unless $2a = 0$ for all $a \in A$, so ignore those kinds of abelian groups).  In the semidirect product $A \rtimes \mathbf Z/(2)$ where
$$
(a,b)(c,d) = (a + (-1)^bc,b+d), 
$$
we have
$$
(0,1)(a,0)(0,1)^{-1} = (-a,0).
$$
Even though $a \mapsto -a$ is an automorphism of $A$ that is not  conjugation on $A$ by some element of $A$ (because $A$ is abelian and we assume $-a \not= a$ for some $a$), by viewing $A$ as a subgroup of the semidirect product $A \rtimes \mathbf Z/(2)$ we can realize the automorphism $a \mapsto -a$ of $A$ as conjugation by some element in a larger group that has $A$ naturally embedded as a subgroup.
Example.  For a nontrivial group $G$, consider $G \times G$, a direct product of $G$ with itself. It has the swap automorphism $s$ where $s(g,g') = (g',g)$. This is an automorphism of order $2$.  In the semidirect product $(G \times G) \rtimes \mathbf Z/(2)$ where
$$
((g_1,g_1'),b)((g_2,g_2'),d) = ((g_1,g_1')s^b(g_2,g_2'),b+d), 
$$
we have
$$
((e,e),1)((g,g'),0)((e,e),1)^{-1} = (s(g,g'),0) = ((g',g),0),  
$$
so by viewing $G \times G$ as a subgroup of $(G \times G) \rtimes \mathbf Z/(2)$, the swap automorphism $(g,g') \mapsto (g',g)$ of $G \times G$ is conjugation on $G \times G$ by an element of a larger group containing $G \times G$ as a subgroup.
Example.  Let's collect together all automorphisms of a group $N$: set $H = {\rm Aut}(N)$ and let $\varphi \colon H \to {\rm Aut}(N)$ be the identity function: $\varphi_f = f$ for all $f \in {\rm Aut}(N)$. The group $N \rtimes_\varphi {\rm Aut}(N)$ contains $N$ as a subgroup, and for all $f$ in ${\rm Aut}(N)$ and $n \in N$,
$$
(1,f)(n,1)(1,f)^{-1} = (\varphi_f(n),1) = (f(n),1).
$$
Thus every automorphism of a group $N$ can be regarded as a conjugation on $N$ from within a larger group.
A: If $(A,+)$ is an abelian group then the only group conjugation is the identity, but $a\mapsto -a$ is an automorphism, and it is only the identity if every element of $A$ has $a+a=0.$
If $(A,+)$ is abelian and finite, and $\gcd(k,|A|)=1,$ you also get the automorphism $a\mapsto ka.$
To give a non-abelian example, if $G_1$ is non-abelian, and $G=G_1\times G_1$ then the map $(g_1,g_2)\mapsto (g_2,g_1)$ is an automorphism of $G$ which is not a conjugacy. That's because any conjugate automorphism sends $(1,g)\mapsto (1,hgh^{-1}),$ and in particular, cannot send $(1,g)\mapsto (g,1)$ when $g\neq 1.$
