it depends on "what a homomorphism is". since you appear to be talking about vector spaces, in that context a homomorphism is a linear mapping. usually "homomorphic to" in this context means:

there is a linear map L:V-->W with im(L) = W.

this is clearly reflexive: V~V since we can take L(v) = v.

it is not symmetric: we always have V~0, since we can take L(v) = 0. however, if dim(V) > 0 there is NO linear map L:{0}-->V, since no matter how we define L, im(L) = {0}, so there are no basis vectors at all in im(L), and V certainly has some (since dim(V) > 0).

it is transitive: suppose U~V, and V~W. so we have linear maps L:U-->V and M:V-->W that are onto. given any w in W, there is some v in V with M(v) = w, since M is onto. given any v in M (including the one (or more) with M(v) = w), we have some u in U with L(u) = v.

consider the map M∘L:U-->W. for any w in W, pick u so that L(u) = v, and M(v) = w (as above).

then M∘L(u) = M(L(u)) = M(v) = w. this makes sense because im(L) = dom(M).

if "homomorphic to" merely means a homomorphism L:V-->W exists, then:

V~V (we can still use L(v) = v).

V~W=> W~V (we can always use the 0-map from W to V).

U~V & V~W => U~W (we can always use the 0-map from U to W).

what makes the first definition trivial is that V~W for ANY choice of V and W. all it really says is "any vector space V has a 0-subspace, {0}", which does not allows us to distinguish between vector spaces at all (the equivalence class of V is EVERY vector space, over ANY field).

a little thought shows that to keep "homomorphic to" as some NON-trivial equivalence, we need to fix symmetry, that is: we need to have an "inverse homomorphism". what do we call onto homomorphisms that have inverses?