We were given the following question:
Let $\phi : G \rightarrow G$ be an automorphism of the finite group $G$ which fixes only the identity element. Show that every element of G can be written in the form $g^{-1}\phi(g)$. Moreover, show that if $\phi \circ \phi = id_{G}$, then G is an abelian group and the group is of odd order.
For the first part of the proof, here is my reasoning, though I do not know if it's correct, or how I can prove it. Since G is a group, and we define an automorphism, essentially it is a permutation of all the elements. So for any $g \in G$, we can find $\phi(g)$ which maps it to g. But this is not a formal proof. And my reasoning feels a little shaky.
For the second part, if we can prove the first part, we have that $\forall$ $a,b \in G$, we have that $a \times b = a^{-1}\phi(a) \times b^{-1}\phi(b)$. But how can we leverage here the fact that $\phi \circ \phi = id_{G}$? We don't have $\phi \circ \phi$ anywhere in the equation unless I am understanding something incorrectly. So how can I show the group is abelian of odd order?