Why is $H \rtimes_{\varphi} K$ isomorphic to $H \rtimes_{\varphi \circ \phi} K$ where $\phi \in \text{Aut}(K)$?
I can see that $\varphi(K) = \varphi(\phi(K))$, but it is not clear to me how the semidirect products themselves are isomorphic. I tried constructing an isomorphism, but they were really messy and didn't seem to go anywhere.
Is there any easy way to see why this is true?
I'm going to use the usual notation, and write your groups as $H \rtimes_{\varphi} K$ (the normal subgroup doesn't get the extra bar on the product sign).
Let's write the general element of $H \rtimes_{\varphi} K$ as $kh$, where the product is given by $$ (k_1h_1)(k_2h_2) = (k_1k_2)(h_1^{\varphi(k_2)}h_2) $$
You can then define a map $f:H \rtimes_{\varphi} K\rightarrow H \rtimes_{\varphi \circ \phi} K$ $$ f(kh) = \phi^{-1}(k)h$$
This is clearly a bijection as a set map, so it just remains to show it's a homomorphism. We have \begin{align} f(k_1h_1k_2h_2) &= f(k_1k_2h_1^{\varphi(k_2)}h_2)\\ &= \phi^{-1}(k_1k_2)h_1^{\varphi(k_2)}h_2\\ &= \phi^{-1}(k_1)\phi^{-1}(k_2)h_1^{\varphi(k_2)}h_2 \end{align}
Meanwhile, in $H \rtimes_{\varphi \circ \phi} K$, \begin{align} f(k_1h_1)f(k_2h_2) &= \phi^{-1}(k)h_1\phi^{-1}(k)h_2\\ &= \phi^{-1}(k_1)\phi^{-1}(k_2)h_1^{\varphi \circ \phi\circ \phi^{-1}(k_2)}h_2\\ &= \phi^{-1}(k_1)\phi^{-1}(k_2)h_1^{\varphi(k_2)}h_2 \end{align}
which agrees with the above. So this is an isomorphism.