This question already has an answer here:
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?