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There is a similar discussion here:
Is any homomorphism between two isomorphic fields an isomorphism?

What I want to assume further is:

Suppose there are two isomorphic groups $\mathcal{M}$, $\mathcal{N}$ with operator multiplication, i.e., there exists an isomorphism between two sets, $\phi: \mathcal{M} \mapsto \mathcal{N}$.

Now suppose that I have another homomorphism $\psi: \mathcal{M} \mapsto \mathcal{N}$ and I know $\psi$ is onto does this guarantee that $\psi$ is an isomorphism?

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    What do you mean by "isomorphic sets", or linear maps between sets? Do you mean to say "vector spaces" instead of "sets"? – 2017-01-18
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    Not vector spaces. In my study, $\mathcal{M}$ and $\mathcal{N}$ are groups actually. – 2017-01-18
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    As @EricWofsey was trying to point out, "linear map" only applies to vector spaces or some structure where there is a notion of addition (and sometimes scalar multiplication). A homomorphism is a far more general notion. – 2017-01-18
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    Ok, I try to edit my question. – 2017-01-18
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    If your question is about groups, what does the answer you accepted have anything to do with it?? – 2017-01-18

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As sets, $\mathbb{R}$ is isomorphic to $\mathbb{R}^2 = \mathbb{C}$, but as groups we have $\varphi : \mathbb{C} \to \mathbb{R}$ given by $\varphi(x,y)=x$ is an homomorphism, is onto but not injective.

Edit: The question was changed, but the example still works since $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as groups too.

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    I don't understand how this answers the question at all, which asks for two **groups** that are isomorphic... – 2017-01-18
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    He edited the question. He asked if for two isomorphic sets, which are also groups, every onto homomorphism is an isomorphism. – 2017-01-18
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    Actually the first version of the question did not even mention sets, and the current version explicit starts with "two isomorphic groups". So I don't see why he/she accepted your answer. Probably not even clear about his/her own question! – 2017-01-18
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    Before the edit, it said "Suppose there are two isomorphic sets M, N". – 2017-01-18
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    I'm not trying to nitpick, but there's no such notion as "isomorphic sets" since one needs structure before "isomorphic" even has any meaning. Otherwise it's quite silly to call any two sets isomorphic just because of a bijection. Anyway it's the asker's problem now so I don't care. =) – 2017-01-18
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    You're right, the correct name would be equipotent sets. I just think the asker did not know the correct word. Still, I edited it. – 2017-01-18
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    Sure. Perhaps note that the reals and complex numbers are isomorphic as groups under addition. Interestingly it seems the proof of that requires something like a Hamel basis. I was wondering whether I missed some simple trick, but apparently not (http://math.stackexchange.com/q/1114241/21820). – 2017-01-18