$G\rtimes_\phi H \cong G\rtimes_\psi H$ when certain automorphisms exist

I think that eventually you have to get down to defining the map explicitly and checking that it is an isomorphism.

But, conceptually: in general, an automorphism of $H$ will not induce an automorphism of $G\rtimes_{\phi} H$, because the action of $f(h)$ need not be related to the action of $h$ (of course, you get a different semidirect product, $G\rtimes_{\phi\circ f}H$, which may or may not be isomorphic). The automorphism $a$ provides the necessary "correction" by having $f(h)$ act on $a^{-1}(g)$ (instead of on $g$), so that the action of $f(h)$ on $a^{-1}(g)$ agrees with the original action of $h$ on $g$.

Intuitively: $H$ used to act on $G$ "in English", but after applying $f$ it is now trying to act "in Russian"; $a$ translates from Russian to English, so $(a^{-1}(g))^{f(h)}$ now makes sense, because $a^{-1}(g)$ is "in Russian", which $f(h)$ understands; then you translate the result back into English by applying $a$. It works exactly like a change-of-basis automorphism works for linear transformations. Interpreted that way, $(g,h)\mapsto (a(g),f(h))$ makes sense (that is, seems like the obvious thing to try), because we are simply performing the translations in both coordinates.