Let me add something to countinghaus' fine answer.
You also asked "why g is an automorphism", something that has not been satisfactorily answered yet.
It is in fact true in much bigger generality. Recall that one axiom of a group action is that for all $x \in X$ and $g,h \in G$ we have
$$
g \cdot_{\text{a}} (h\cdot_{\text{a}} x) = (g \cdot_{\text{g}} h) \cdot_{\text{a}} x
$$
where i labeled the group multiplication with g and the action operation with a.
Applying this to $h = g^{-1}$ and using that the unit element of the group acts as the identity on $X$, we see that we have found an inverse for our homomorphic map given by $g$. Hence it is an automorphism.
Note that one needs to be precise. Of course $g^{-1}$ is the inverse of $g$ in the group, that we all know. However, we've just shown something else: that the holomorphic map given by $g^{-1}$ is an inverse of the holomorphic map given by $g$.
Lastly, it works in any category $\mathcal{C}$. If we say some group acts on some object in $\mathcal{C}$, we normally require that the elements of the group induce morphisms in $\mathcal{C}$, which have a two sided inverse by the reasoning above. Hence they are isomorphisms in $\mathcal{C}$.
Hope that helps.