Showing the existence of an automorphism between two semidirect products Let $H$ and $K$ be two groups, and let $\phi$ and $\phi’$ be two group homomorphisms from $H$ to $Aut(K)$. Then it can shown that if there’s an automorphism $\sigma$ in $H$ such that $\phi’=\phi \circ \sigma$, then the groups given by the two semidirect products $K \rtimes_{\phi}H$ and $K \rtimes_{\phi’}H$ are isomorphic.
According to my lecture notes, to do so we can check that the map $\rho: K \rtimes_{\phi}H \rightarrow K \rtimes_{\phi’}H$ such that $\rho(k,h)=(k,\sigma^{-1}(h))$ is an homomorphism (it’s clearly a bijective map since $\sigma$ is a group isomorphism).
What I don’t see so clearly is, how did they come up with such map. I would have thought $h\in H$ should be mapped to another element in $H$ related to $h$ by $\sigma$, but I don’t understand why it’s $\sigma^{-1}(h)$ instead of $\sigma(h)$. Where does the inverse come from?
Also (probably unrelated question but…), would this map $\rho$ be a map between (split) extensions as well?
 A: The reason you use $\sigma^{-1}$ rather than $\sigma$ is because you are trying to define a kind of "change-of-basis" or "translation" morphism.  Those often involve the "inverse" map, because you want things to happen as they do in the domain, when you are in the codomain.
Note that $\phi'=\phi\circ\sigma$. Consider what happens with $(e,h)$ multiplying $(k,e)$. In the first group, you have $(\phi(h)(k),h)$. So you want $\rho(h)$ to act on $k$ as $\phi(h)$ does; but instead $\rho(h)$ will act as $\phi'(\rho(h))$. Since $\phi'=\phi\circ\sigma$, that means that you will get $\phi(\sigma(\rho(h))$. In order to ensure that this is $\phi(h)$, you need $\sigma(\rho(h))=h$, so you need $\rho(h)=\sigma^{-1}(h)$. Thus, you need to use $\sigma^{-1}$ instead of $\sigma$.
For your second question: if the extension is a split extension, then $\phi$ is trivial, so the existence of $\sigma$ guarantees that $\phi'$ is also trivial (since $\phi$ is the map sending everything to the identity). The map will work, because $\sigma$ will just act like an automorphism of the $H$ summand.
