What does the endomorphism group of an object tell us about the object in question? For example:
What conclusions can be drawn about the relations between two objects with the same group of endomorphism?
Can we tell from End(A) if A is Abelian or not?
Does End(A) contain information about the sub-objects of A?
Any information or references to information about this is highly appreciated.
 A: For abelian groups, the ring End(A) is very important.  As far as non-abelian groups A go, End(A) is not even (usually considered) a group.
"Adding" homomorphisms doesn't work in the non-abelian case.
If you define (f+g)(x) = f(x) + g(x), then (f+g)(x+y) = f(x+y) + g(x+y) = f(x) + f(y) + g(x) + g(y), but (f+g)(x) + (f+g)(y) = f(x) + g(x) + f(y) + g(y).  To conclude that:
    f(y) + g(x) = g(x) + f(y)
are equal, you use that + is commutative, that A is abelian.  More precisely, if you take f=g to be the identity endomorphism, then f+g is an endomorphism iff A is abelian.
"Composing" homomorphisms doesn't work to form a group, since they are not invertible.
Aut(A), the group of invertible endomorphisms, does form a group.  Aut(A) does not determine if a group is abelian or not: 4×2 and the dihedral group of order 8 have isomorphic automorphism groups.
Instead of a ring, End(A) sits inside the "near-ring" of self-maps.  See the wikipedia article on nearring for an explanation.
A: Here is a partial answer (although it uses the ring structure on the set of endomorphisms). A torsion abelian group is cyclic if and only if any two elements in $End(A)$ commute with respect to composition (in other words, $End(A)$ is a commutative ring).
http://www.springerlink.com/content/m72222448q6j7327/
There is a whole book on this subject: Endomorphism rings of abelian groups.
