Semidirect product: general automorphism always results in a conjugation

Use the same idea. Observe

$$\begin{array}{ll} (n_1,h_1)(n_2,h_2) & =(n_1,e_H)(e_N,h_1)(n_2,e_H)(e_N,h_2) \\ & = (n_1,e_H)\color{Blue}{(e_N,h_1)(n_2,e_H)(e_N,h_1^{-1})}(e_N,h_1,)(e_N,h_2) \end{array} $$

and

$$\begin{array}{ll} (n_1,h_1)(n_2,h_2) & =(n_1\phi_{h_1}(n_2),h_1h_2) \\ & = (n_1,e_H)\color{Blue}{(\phi_{h_1}(n_2),e_H)}(e_N,h_1)(e_N,h_2). \end{array} $$

This means when we conjugate elements of $N\times\{e_H\}$ by elements of $\{e_N\}\times H$, we get the same thing as if we apply the elements of $H$ as automorphisms to $N$, then put it in $N\times\{e_H\}$.

Tuples are annoying and obfuscate the algebra in my opinion though. We should think of the semidirect product $N\rtimes H$ as the free product $N*H$ (whose elements are words formed from using the elements of $N$ and $H$ as letters) modulo the relation that conjugating elements of $N$ by elements of $H$ yields the same thing as if we applied the corresponding automorphism.

That is, elements of $N\rtimes H$ look like words formed from elements of $N$ and $H$. Their identity elements are identified as the same group element in $N\rtimes H$. Elements of $N$ multiply among themselves as usual, and same for elements of $H$ multiplying among themselves. But every instance of the word $hnh^{-1}$ ($h\in H,n\in N$) may be simplified to $\phi_h(n)$, and that is the only relation imposed on multiplication between elements of the two subgroups $N$ and $H$.

Using this definition, it's easy to see that $hn=(hnh^{-1})h=\phi_h(n)h$ so $HN=NH$ within $N\rtimes H$, and every element of $H$ can be "slid past" an element of $N$ to the right (although it changes the element of $N$ along the way). As a result, every word $\cdots h_{-1}n_{-1}h_0n_0h_1n_1\cdots$ (finitely many letters of course) can be simplified via this sliding rule to the canonical form $nh$.

Writing $n_1h_1=n_2h_2$ yields $h_1h_2^{-1}=n_1^{-1}n_2$, but the only element in $N\cap H$ (when we treat $N,H$ as subgroups of $N\rtimes H$) is the identity, so $h_1=h_2$ and $n_1=n_2$. Thus $N\rtimes H$ can be bijected with $N\times H$ set-theoretically. In order to transport the multiplication over, it remains to see how $(n_1h_1)(n_2h_2)$ simplifies to $n_3h_3$, which is something you've essentially already done.