Given a rational number a/b does there exist a finite group G and an automorphism f s.t. f maps exactly a/b elements of G to their own inverses?
I was helping a friend prepare for his intro abstract final and he mentioned the professor had once asked the question name a group and an automorphism that takes 3/4 of the elements of the group to their own inverses (D4, identity). I tried to figure out how to approach this question in general but can't see how.
Can we construct the group?
Does anything change if we allow infinite groups?