Question of Automorphism $T$ on Finite group with property that $T(x)=x$ only for $x=e$ 
Let $G$ be finite group, $T$ be automorphism on $G$ with property that $T(x)=x$ only for $x=e$. Then 
  1) every $g \in G $ can be written as $g=T(x)x^{-1}$ for some $x\in G$
  2) Furthermore if $T^2=\mathrm{Identity}$ then $G$ is abelian group

My first question is the following:

0) What is so important about condition $T(x)=x$ only for $x=e$? Is there any significance associated with it?

My Attempt for 1:

Let $G =\{x_1,x_2,\dotsc,x_n\}$ be finite  group. After applying required transformation we get $\{T(x_1)x_1^{-1},T(x_2)x_2^{-1},\dotsc,T(x_n)x_n^{-1}\}$. Now on contary assume they are not distanst.
  $T(x_i)x_i^{-1}=T(x_j)x_j^{-1}$ now i.e. $T(x_j)^{-1}T(x_i)=x_j^{-1}x_i$ which leads $T(x_j^{-1}x_i)=x_j^{-1}x_i$ due to given condition $x_j^{-1}x_i=e$ that means contradiction. Hence all are distanct. By the pigenhole priciple we get required.

Is I am giving right argument?
2) Second question Here I am not able to use given data.
Any Help will be appreciated.

Since $G$ is finite, the map $f\colon G\to G$ defined by $x\mapsto f(x):=T(x)x^{-1}$ is surjective (so the property 1) holds) if and only if it is injective. And it is injective, because:
\begin{alignat}{1}
&f(x)=f(y)&&\Longrightarrow \\
&T(x)x^{-1}=T(y)y^{-1}&&\Longrightarrow \\
&T(y^{-1}x)=y^{-1}x&&\Longrightarrow \\
&y^{-1}x=e&&\Longrightarrow \\
&x=y&&\Longrightarrow \\
\end{alignat}
where the middle implication holds precisely by the assumption $T(g)=g\Longrightarrow g=e$.
For the second question, note that if, in addition, $T^2=Id$, then $\forall g \in G,\exists x\in G$ such that:
\begin{alignat}{1}
T(g) &= T(T(x)x^{-1}) \\
&= T(T(x))T(x^{-1}) \\
&= xT(x)^{-1} \\
&= (T(x)x^{-1})^{-1} \\
&= g^{-1}
\end{alignat}
Therefore, for every $x,y\in G$, both:
$$T(xy)=T(x)T(y)=x^{-1}y^{-1}=(yx)^{-1}$$ and: $$T(xy)=(xy)^{-1}$$
whence $yx=xy$.
