Why is $H \rtimes_{\varphi} K$ isomorphic to $H \rtimes_{\varphi \circ \phi} K$ where $\phi \in \text{Aut}(K)$? 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?
 A: 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.
