Inner semidirect product as outer semidirect Let $G$ be an (ambient) group which splits into an (inner) semidirect product $G_1\rtimes_\varphi G_2$ of two subgroups (satisfying other conditions). We can define $\varphi:G_2\to\mathrm{Aut}(G_1),\varphi(g_2)(g_1)=g_2g_1g_2^{-1}$.
It seems that only inner automorphisms, i.e., of the form $g(\cdot)g^{-1}$, can be used to build the original group. But the group $\mathrm{Aut}(G_1)$ can be quite messy (with a lot of outer automorphism).

Can we construct the $\varphi$ not only with those inner automorphisms?

 A: For an outer semi-direct product you just need arbitrary groups $H$ and $N$ and an arbitrary homomorphism $\theta\colon H\to\operatorname{Aut}(N), h\mapsto \theta_h$. The image of $\theta$ does not have to be contained in the inner automorphisms $\operatorname{Inn}(N)\le\operatorname{Aut}(N)$.
However, in the semidirect product $G=N\rtimes_\theta H$, the automorphisms of $N$ in the image of $\theta$ are realized as elements of $\operatorname{Inn}(G)$: the product is defined as
$$
(n,h)(n',h') = (n\,\theta_h(n'), hh')
$$
which yields inverses
$$
(n,h)^{-1} = (\theta_{h^{-1}}(n^{-1}), h^{-1}).
$$
Denoting by $\iota_H\colon H\to G$ and $\iota_N\colon N\to G$ the embeddings $h\mapsto(e_N, h)$ and $n\mapsto(n, e_H)$ respectively, this yields
\begin{align*}
\iota_H(h) \iota_N(n) \iota_H(h)^{-1} &=
(e_N,h)(n,e_H)(e_N,h^{-1}) \\&=
(\theta_h(n),h)(e_N,h^{-1}) =
(\theta_h(n), e_H) = \iota_N(\theta_h(n)).
\end{align*}
Hence, the subgroup $\operatorname{Im}(\theta)\le\operatorname{Aut}(N)$, which might contain non-inner automorphisms of $N$, gets mapped into the inner automorphism group $\operatorname{Inn}(G)$ of $G$ by sending $\theta_h$ to conjugation with $\iota_H(h)$ in the semidirect product.
