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 $N \rtimes_\varphi H$ 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.
