I came across this question while working on a problem from my practice final for abstract algebra.
I've set up the groups, and proved the homomorphism, and shown that the two groups have the same order. So can I state that all homomorphic groups of the same order are isomorphic?
Here's what I have so far.
Here is a trivial counter-example.
Let $G$ and $H$ be two groups of the same order, but not isomorphic. For any pair of groups, there is always the $0$ homomorphism between them, i.e. the map $0: G \to H: G \ni g \mapsto e \in H$. This is a homomorphism trivially, and so there is a homomorphism from $G$ to $H$, but somehow it is not a useful homomorphism. It's not even close to an isomorphism.
I expect the relationship you're after is given by the First Isomorphism Theorem, which in a sense tells you how a homomorphism can be 'close' to an isommorphism. In particular, if $\phi: G \to H$ is a homomorphism, this theorem asserts that $G/\ker \phi \simeq H$.
Edit: If by homomorphic you mean that $G$ has a homomorphic image which is ONTO $H$, then yes your claim is true, since for groups with the same finite order, being surjective implies being injective.