If an automorphism sends more than half of the elements to themselves then it's the identity

Question

Suppose $G$ is a finite group and $f$ is an automorphism which sends more than half of the elements in $G$ to back to them, meaning $f(a)=a$. Conclude that $f$ is the identity.

I know it's something with symmetry but I failed to understand how to prove it formally.

Thanks!

Answer 1

The set of all elements fixed by the automorphism is subgroup of $G$. If the order of this subgroup is more than half the size of $G$, then its order cannot divide the order of $G$ unless it is $G$ itself.