Caveat: I am new to abstract algebra.
I can't think of a slick proof. Essentially I show this case by case.
Case 1:
Because the order of G is 1, then the only element of G is the identity element, G={1}. Further more the only automorphic map
is then the identity map.
Case 2:
Because the order of G is 2, then the only two elements of G are the identity element and some other element, G={1,a}. Because a homomorphism must contain all the inverses for its elements,
must be its own inverse,
. Furthermore, because a homomorphism maps identities to identities and inverses to inverses the only possible homomorphism (and therefore automorphism) is again the identity map.
Case 3:
We need to think up an automorphism that is not the identity map. We need to show that this map is an automorphism. That is as far as I got.
Sorry,
hope this helps some