To show a composition of two maps, $f, g$ is an automorphism I need to show they are bijective and homomorphism. The homomorphism part is easy: If we have $f, g: G \rightarrow G$, then for some element $a,b \in G$ we get

$$ f \circ g (ab) = f(g(ab)) = f(g(a)g(b))$$

as $g4$ is an isomorphism. Then as $f$ is an isomorphism, we also get

$$f(g(ab)) = f \circ g(a) f \circ g(b)$$

I'm pretty sure that's correct. 

Now to prove for bijection, I first thought about checking to see if the kernel was trivial as that would show injection, $g(a,b) = (0,0)$ iff $a = b = 0$. So putting this in the composition gives us $f \circ g(0,0) = f(g(0,0)) = f(0,0) = (0,0)$, hence the map is injective. We see from the definition of the map that every element in it maps to some element in $G$ and so the map is also surjective and therefore bijective. 

Is that correct?