$f$ be a homomorphism from $G$ onto $G'$. I know that $f$ is an $n$-to-one function where $n=|\operatorname{Ker} f|$. So to each element in the homomorphic image $G'=f(G)$ there corresponds exactly $n$ elements in $G$. So $|G|=n*|G'|$. Is this right?
If so, does that mean if $|G|=|G'|$, then $f$ is an isomorphism?
Yes. This is a consequence of the first isomorphism theorem. Given a homomorphism $f: G \to L$ we have an induced iso $G/\mathrm{ker}(f) \cong \mathrm{im}(f)$. In the event everyone's finite you can say that the order of $G/\mathrm{ker}(f)$ is $|G|/|\mathrm{ker}(f)| = |\mathrm{im}(f)|$ which gives you everything you need
