Let $N,H$ be groups, $\phi \colon H\rightarrow Aut(N)$ be a homomorphism, $f\in Aut(N)$ and $\hat{f}$ be an inner automorphism of $Aut(N)$ induced by $f$. Then $N\rtimes_{\phi} H \cong N\rtimes_{\hat{f}\circ\phi}H$. Also, if $\psi \in Aut(H)$, then $N\rtimes_{\phi}H\cong N\rtimes_{\psi\circ\phi}H$. What are sufficient conditions for isomorphism of semidirect products? Are there any other criterias for isomorphism of semidirect products?

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 $G = \langle x,y \mid x^{29}= y^{29}=z^7=1, xy=yx, x^z=x^7, y^z=y^{16} \rangle,$ 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! 


Aside from the two ways you've mentioned, I think there are just two other ways of getting an isomorphism (i.e. any isomorphism of semidirect products with same $N$ and $H$ is some combination of all four):


