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.
 A: You look like you've got a good idea for part $1,$ though it's hard to read and understand. Here's how I'd adjust it:

Suppose $G$ is a finite group of order $n,$ so that $G =\{x_1,x_2,\dots,x_n\},$ where the $x_i$ are distinct. After applying the transformation $x\mapsto T(x)x^{-1},$ we get $\{T(x_1)x_1^{-1},T(x_2)x_2^{-1},\dots,T(x_n)x_n^{-1}\}.$ We want to show that this set is equal to $G.$ We know that it is a subset of $G$ since $G$ is a group, and if we can show that its elements are distinct, then it will be equal to $G$ by Pigeonhole principle. On the contrary, suppose that the elements are not all distinct, so that for some $i,j$ with $i\ne j,$ we have $$T(x_i)x_i^{-1}=T(x_j)x_j^{-1}.$$ Then $$T(x_j)^{-1}T(x_i)=x_j^{-1}x_i,$$ which leads to $$T(x_j^{-1}x_i)=x_j^{-1}x_i$$ by homomorphism properties, and so $x_j^{-1}x_i=e$ by the assumed property of $T$. But then $x_i=x_j,$ contradicting the fact that $i,j$ (and so $x_i,x_j$) are distinct.

Basically, it's the same thing you were saying, but with improved formatting, phrasing/spelling. I also removed the undesirable assumption (from the original post) that $G$ was abelian. Moreover, I added the assumption that $G$ had order $n.$ (Do you see why that's necessary?) Also, do you see why the condition $T(x)=x\implies x=e$ was important, here?
Edit: It's worth noting, though, that we can do even better.
Consider any group $G$ and any automorphism $T:G\to G,$ and define a map $f$ by $f(x):=T(x)x^{-1}$ for all $x\in G.$ I claim that the following statements are equivalent:

*

*$T(x)=x$ only if $x=e.$

*$f$ is a one-to-one map of $G$ into $G.$
Since $T(x)\in G$ and $x^{-1}\in G$ for all $x\in G$ by automorphism and group properties, then we automatically have that $f$ is a map from $G$ into $G$ regardless of any other assumptions.
Now, if $T$ has the property that $T(x)=x$ only if $x=e,$ then taking any $x,y\in G$ such that $f(x)=f(y),$ we have by definition that $T(x)x^{-1}=T(y)y^{-1},$ so $T(y)^{-1}T(x)=y^{-1}x,$ and so $T\left(y^{-1}x\right)=y^{-1}x$ by homomorphism properties. By assumption, it follows that $y^{-1}x=e,$ so $x=y,$ meaning that $f$ is a one-to-one map on $G$.
On the other hand, if we suppose that $f$ is a one-to-one map on $G,$ then take any $x\in G$ and suppose that $T(x)=x.$ Thus, $T(x)x^{-1}=e,$ or equivalently, $f(x)=e,$ but noting that $f(e):=T(e)e^{-1},$ then $f(e)=ee^{-1}$ by homomorphism properties, so $f(e)=e=f(x).$ Since $f$ was assumed to be one-to-one, it follows that $x=e.$ Thus, $T(x)=x$ only if $x=e.$ $\Box$
As an immediate corrolary, the following are equivalent for any finite group $G$ and automorphism $T:G\to G.$

*

*$T(x)=x$ only if $x=e.$

*$x\mapsto T(x)x^{-1}$ is a one-to-one mapping of $G$ onto $G.$

As for part $2,$ let me give you a hint.
The following are equivalent: $$T^2=id\\\forall x\in G,T\bigl(T(x)\bigr)=x\tag{$\star$}$$
The following are also equivalent: $$G\textrm{ is abelian}\\\forall g,h\in G,gh=hg\\\forall g,h\in G,ghg^{-1}h^{-1}=e\\\forall g,h\in G,(gh)^{-1}=g^{-1}h^{-1}\tag{$\heartsuit$}$$
Now, using part 1, we know that for each $g\in G,$ there is an $x\in G$ such that $g=T(x)x^{-1}$--let's call this by the name $x_g,$ so we know which $g$ it corresponds to.
Applying $(\star)$ and part $1$ then gives us $$T(g)=T\bigl(T(x_g)x_g^{-1}\bigr)=T\bigl(T(x_g)\bigr)T(x_g)^{-1}=x_gT(x_g)^{-1}=\bigl(T(x_g)x_g^{-1}\bigr)^{-1}=g^{-1}$$ for each $g\in G.$ Can you use this together with $(\heartsuit)$ to show that $G$ is abelian?
A: 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 $yx=xy$.
