Finding a different proof.

Question

There's a problem in topics in algebra by I.N.Herstein in the chapter of automorphism. If |G|$\geq$3(yet finite), then |Aut(G)|$\gneqq$1. Now if G is non abelian, then Z(G)$\neq$G. Thus there exists an inner automorphism which is not the identity mapping. Now, if G is abelian, then T:G$\rightarrow$G, T(g) = -g. Then T is a non trivial automorphism, iff every element of G is not of order 2. Thus the case that remains is when every element of G is of order 2. Now I can prove it using little linear algebra and direct product. But how can I get it done using only the concept of automorphism, and the isomorphism theorems?