Let $N,H$ be groups, $\phi \colon H\rightarrow Aut(N)$ be a homomorphism, if $\psi \in Aut(H)$, prove that $$N\rtimes_{\phi}H\cong N\rtimes_{\phi\circ\psi}H.$$
This is mentioned in the original post. I am seeking proof of the statement.
Edit: I have changed the order of $\phi$ and $\psi$ and the composition acts from right to left. Sorry about the vagueness.
Define $\varphi: N\rtimes_\phi H\to N\rtimes_{\phi\circ\psi}H$ by
$$ (n,h)\mapsto (n,\psi^{-1}(h)) .$$ So it suffices to show that $\varphi$ is a group homomorphism.
\begin{align}
\varphi((n_1,h_1)\cdot_\phi(n_2,h_2))&=\varphi((n_1\phi(h_1)(n_2),h_1h_2))\\
&=(n_1\phi(h_1)(n_2),\psi^{-1}(h_1)\psi^{-1}(h_2))
\end{align}
On the other hand,
\begin{align}
\varphi((n_1,h_1))\cdot_{\phi\circ\psi}\varphi((n_2,h_2))&=(n_1,\psi^{-1}(h_1))\cdot_{\phi\circ\psi}(n_2,\psi^{-1}(h_2))\\
&=(n_1\phi\circ\psi\circ\psi^{-1}(h_1)(n_2),\psi^{-1}(h_1)\psi^{-1}(h_2))
\end{align}
Best Answer
$\DeclareMathOperator{\Aut}{Aut}$$\renewcommand{\phi}{\varphi}$I apologise for using a different notation (composition left-to-right and homo/auto-morphisms written as exponents), but if I do it differently, I might easily do it wrong. May try and translate it to different notation if deemed useful.
In $H \ltimes_{\phi} N$ the operation is given by $$ (h_{1}, n_{1}) \cdot (h_{2}, n_{2}) = (h_{1} h_{2}, n_{1}^{h_{2}^{\phi}} n_{2}). $$ (So in my notation $h_{2}^{\phi} \in \Aut(N)$ is the image of $h_{2}$ under $\phi$, and $n_{1}^{h_{2}^{\phi}}$ denotes its operation on $n_{1}$.)
In $H \ltimes_{\psi \phi} N$ the operation is given by $$ (h_{1}, n_{1}) \circ (h_{2}, n_{2}) = (h_{1} h_{2}, n_{1}^{h_{2}^{\psi \phi}} n_{2}). $$ (Here $\psi \phi$ is the composition of $\psi$ first and $\phi$ second.)
Now an isomorphism between the two groups is given by \begin{align} G :\ &H \ltimes_{\phi} N \to H \ltimes_{\psi \phi} N\\ &(h, n) \mapsto (h^{\psi^{-1}}, n). \end{align} (Note the critical inverse, see the calculation below.)
In fact
$$ G((h_{1}, n_{1})) \circ G((h_{2}, n_{2})) = (h_{1}^{\psi^{-1}}, n_{1}) \circ (h_{2}^{\psi^{-1}}, n_{2}) = (h_{1}^{\psi^{-1}} h_{2}^{\psi^{-1}}, n_{1}^{h_{2}^{\psi^{-1} \psi \phi}} n_{2}) = ((h_{1} h_{2})^{\psi^{-1}}, n_{1}^{h_{2}^{\phi}} n_{2}), $$ which equals $$ G((h_{1}, n_{1}) \cdot (h_{2}, n_{2})) = G((h_{1} h_{2}, n_{1}^{h_{2}^{\phi}} n_{2})) = ((h_{1} h_{2})^{\psi^{-1}}, n_{1}^{h_{2}^{\phi}} n_{2}). $$