I will assume you mean abelian group endomorphisms, and not ring endomorphisms. I will write Z_{4}additively.

As a group, Z_{4}is generated by 1, so any endomorphism is completely determined by the image of 1. We have 4 choices for such an endomorphism f:

f(1) = 0. This is the trivial map, which sends everything to 0.

f(1) = 1. This is the identity map, which is an automorphism.

f(1) = 2. This is a quotient map onto the subgroup {0,2} (which sends a to 2a (mod 4)).

f(1) = 3. This is another automorphism, which sends a to -a.

So Hom(Z_{4},Z_{4}) has 4 elements, and in fact is isomorphic to the multiplicative monoid of Z_{4}.

By the way, this is NOT a group, only the automorphisms Aut(Z_{4}) form a group, of order 2. The endomorphisms: a-->0, a --> 2a, are not invertible.