This is not an answer to your question, but there is another way that an isomorphism can arise. It is possible for two semidirect products $N_1 \rtimes H_1$ and $N_2 \rtimes H_2$ (with $N_1 \cong N_2$, $H_1 \cong H_2$) to be isomorphic as groups, but for there to be no isomorphism that maps $N_1$ to $N_2$. An example of this is the group

$G = \langle x,y,z \mid x^{29}= y^{29}=z^7=1, xy=yx, x^z=x^7, y^z=y^{16} \rangle,$
which is an extension of $C_{29} \times C_{29}$ by $C_7$. Let $N_1$ and $N_2$ be the normal subgroups of order 29 generated by $x$ and $y$. Then $G/N_1 \cong G/N_2$ is the unique nonabelian group of order $29 \times 7$, but there is no automorphism of $G$ that maps $N_1$ to $N_2$, so this group can be expressed as a semidirect product of $C_{29}$ by the nonabelian group of order $29 \times 7$ in two different ways.

You might prefer ot restrict your attention to isomorphisms between $N_1 \cong N_2$ and $H_1 \cong H_2$ that map $N_1$ to $N_2$. If you do that, then I don't know the answer to your question in general, but I believe that in the special case when $N$ is an elementary abelian $p$-group, all isomorphisms arise in the ways you have described in your post. I don't feel like trying to write down a proof right now!

criteriasdoes not exist. The plural ofcriterioniscriteria. $\endgroup$ – Chandan Singh Dalawat Feb 5 '11 at 6:292more comments