2.3:

(a) Suppose (x)s=x^-1defines an automorphism of G. This map is always a bijective, so the hypothesis is really just saying (xy)^-1 = x^-1y^-1 for all x,y in G. This implies y^-1x^-1 = x^-1y^-1, which gives x^-1 = y x^-1y^-1, then x^-1y = y x^-1, y = xy x^-1, and finally, yx=xy for all x,y in G. Thus $G$ is abelian.

(b) Suppose s : G ---> G is an automorphism satisfying s² = id and (x)s is different from x for x not equal to e. Consider the function G --> G mapping x to x^-1x^s. We claim this map is injective. Indeed, suppose x^-1x^s = y^-1y^s. Manipulating and using that s is a homomorphism, we get yx^-1 = y^s (x^s)^-1 = (yx^-1)^s. Then yx^-1 = e, by hypothesis, so y=x. Since the map is injective, and G is finite, the map is surjective as well. Then, for any y in G, there is an x in G such that y = x^-1x^s. Applying s, we get that y^s = (x^-1x^s)^s = (x^-1)^sx^ss, which equals (x^s)^-1x since ss = s², and this is equal to (x^-1x^s)^-1, which is y^-1. Thus (y)s = y^-1 for all y in G. Then G is abelian by part a.

 

This shows, for finite groups, the only possible automorphism of order two with no notrivial fixed points is the inversion mapping, in which cae G is abelian. What about infinite groups?