Showing the existence of an automorphism between two semidirect products

Solution 1:

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.