finite group whose only automorphism is identity map Question is to prove that  :
A finite group whose only automorphism is identity map must have order at most $2$.
What i have tried is  :
As any automorphism is trivial, so would be inner automorphism 
i.e., each map for fixed $g\in G $ with $\eta : G\rightarrow G$ taking $h$ to $ghg^{-1}$ is trivial.
Thus, $ghg^{-1}=g$ i.e., $gh=hg$ for all $g\in G$ and $h\in G$ which would say that $G$ is abelian.
So, I would have that $G$ is finite abelian group.
Now, As $G$ is abelian, the map $g\rightarrow g^{-1}$ is an automorphism.
But only automorphism is identity map, so we would have  :
$g=g^{-1}$ i.e., $g^2=e$ for all $g\in G$
So, I would have that $G$ is group with each element of order $2$.
Combining with previous result I would have :

$G$ is a finite abelian group in which each element is of order $2$

I am not able to conclude anything more than this....
A kind of cheating would give something very close :
As group is finite abelian which has each element with order $2$, It should be :
$\mathbb{Z}_2\times \mathbb{Z}_2$ Or
$\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ Or
$\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ Or something very similar to this.
For first group $\mathbb{Z}_2\times \mathbb{Z}_2$  automorphism group is general linear group of order $2$ with entries from $\mathbb{Z}_2$ which is not trivial. So, this should not be the required group.
This would hold for similar cases
So, I feel that i am on right path but i need some help to make it more clear.
Thank you :)
 A: A (finite) abelian group with $g^2=e$ for all $g\in G$ is in fact a (finite dimensional) $\Bbb Z_2$-vector space, and then its automorphisms correspond to invertible matrices over $\Bbb Z_2$, so...
A: You are already very close. 
Indeed, nonabelian groups have nontrivial inner automorphisms, while abelian ones have the automorphism $a\mapsto a^{-1}$, which is trivial only when all $a$ have order $2$.
So we only need to deal with the direct sums of $\mathbb{Z}_2$. But for such groups we have the automorphism \begin{equation}
(x,y)\mapsto (y,x),
\end{equation} which is nontrivial unless we have only one copy in our direct sum.
A: We show that $Aut(G)$ is nontrivial if $G$ is abelian, $|G|>2$ (may be infinite), and all its non-identity elements are of order 2.
Let $G$ be an abelian group with all non-identity elements of order $2$ and $|G|>2$.  It's easy to see that $A = \{1, a_1, a_2, a_1a_2\}$ is a subgroup of $G$, where $a_1$ and $a_2$ are distinct non-identity elements of $G$. Let $B$ be a subgroup of $G$ and also a transversal of the collection of all cosets of $A$. Let $h$ be a nontrivial automorphism on $A$, and we can cronstuct an nontrivial automorphism $f$ on $G$ by defining $f$($b$$a$) = $b$$h$($a$) for all $b \in B$ and $a \in A$. $ So, Aut(G)$ is nontrivial.
