Let $G$ be a finite group, and let $T$ be an automorphism of $G$ such that $T(x) = x$ for $x \in G$ if and only if $x = e$, the identity element in $G$. Then can we represent every element $g$ of $G$ in the form $g = x^{-1} T(x)$ for some $x \in G$?
And if so, then can we also extend this representation so as to represent every $g$ in $G$ as $g = T(y) y^{-1}$ for some element $y \in G$?
Yes, as Derek pointed out.
Let $f(x)=x^{-1}T(x)$, if we show that $f$ is one to one then it must be onto since $G$ is finite.
So let $$f(x)=f(y)$$ $$x^{-1}T(x)=y^{-1}T(y)$$ $$yx^{-1}=T(y)(T(x))^{-1}$$ $$yx^{-1}=T(y)T(x^{-1})=T(yx^{-1}) \implies yx^{-1}=e\implies x=y$$
we are done. Second part of question can be done with same method.