Let G be a group. For g G, conjugation by g, that is, the map x -> gxg^-1, is an automorphism of G. These automorphisms are also called inner automorphisms of the group G.

By a previous theorem, the map x -> x^p is an automorphism of a field of characteristic p.

There is exactly one automorphism of the field Q: the identity.

For an automorphism : Q -> Q, we have (1) = 1, so (2) = (1) + (1) = 2, etc. By induction, (m) = m for positive integers m.

From (0) = 0 it follows that 0 = (m+(-m)) = (m) + (-m) and so (-m) = -m for all positive integers m (here use we that is an automorphism of the additive group of Q).

For a/b Q, with b positive, (a) = (b · a/b) = (b) · (a/b). This implies a = b · (a/b) and so (a/b) = a/b.

If Q is a subfield of the field K and : K -> K is an automorphism, then the same argument shows that (r) = r for each r Q.

Let <g> be a group of order n generated by g. An automorphism is determined by the image of g, which, must be of the form g^j for an integer j with gcd(j,n) = 1. For, otherwise the element g^j does not have the same order as g.

On the other hand, for each such exponent j prime to n the map g -> g^j is an automorphism.