Let G be a group such that Aut(G) is cyclic and not trivial. Then Aut(G) is finite and his order is even.

Since Aut(G) is cyclic then Inn(G) is cyclic so G/Z(G) is cyclic which implies G being abelian. Let us first show that Aut(G) has an element of order 2. Consider the map:

f: x ∈ G |—> x⁻¹ ∈ G

we have (G being abelian)

f(xy)=(xy)⁻¹=y⁻¹x⁻¹=x⁻¹y⁻¹=f(x)f(y)

so f is a group morphism. Now f is trivially surjective and if x⁻¹=y⁻¹ then x=y so f is isomorphism. It follows that Aut(G) has an element of order 2 (note that f cannot be the identity map since otherwise G is elementary abelian of order 2 and so Aut(G)=GL(n,ℤ₂) for some n∈ℕ⁺. But if n=1 Aut(G) is trivial and if n>1 GL(n,ℤ₂) is not cyclic ). Now if by contradiction Aut(G) is infinite then from the classification of cyclic groups Aut(G)=Z then Z would have an element of order 2, that is, there would exist m in Z \ {0} such that m+m=0 absurd. Therefore Aut(G) must be finite and since it has an element of order 2 by Lagrange 2| |Aut(G)| that is |Aut(G)| is even. ■