If $ G $ has no non-trivial automorphism, then $ G $ is abelian and $ g^2 = e $ for all $ g \in G $ . If $ G $ has no non-trivial automorphism, then $ G $ is abelian and $ g^2 = e $ for all $ g \in G $ .
With the assumption, I dont know how to start the proof.
If there is no non-trivial automorphism, then there is only trivial automorpism, the identity morphism. But how can I show  $ g^2 = e $ for all $ g \in G $ with it?
 A: Hint: If $Inn(G)$ is the group of inner isomorphisms, which we are taking to be trivial, apply the following theorem:
$G/Z(G)\cong Inn(G)$. 
Recall: If $G$ is Abelian, what is the relationship between $G$ and $Z(G)$?
A: After the observations above, note that in case $G$ is finite (for infinite groups a similar reasoning applies), it must be an elementary abelian 2-group of rank say $n$. Then $Aut(G) \cong GL(n,2)$. Hence if $|Aut(G)| = 1$ then this can only be the case if $G$ is trivial, or $n = 1$ that is $G \cong C_2$, the cyclic group of order 2.
A: For every elements $a\in G$, consider 
$$\phi_a:G\rightarrow G$$
sending an element $b$ to its conjugate $a^{-1}ba$.
You can easily check that all $\phi_a$ are automorphisms of $G$, called inner automorphisms. By assumption, $\phi_a=id$, which means
$$\phi_a(b)=b$$
for every $a,b\in G$, so that $ba=ab$. 
Alternatively, use
$$\frac{G}{Z(G)}\cong Inn(G)$$
where $Z(G)$ is the center of $G$ and $Inn(G)$ the group of inner automorphisms, which is trivial in your hypothesis.
As for $g^2=e$, you can use abelianity we have just proved. Since $(ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}$, the inversion $g\rightarrow g^{-1}$ is an automorphism, which by assumption is trivial. Multiplying $g=g^{-1}$ by $g$ gives $g^2=e$
