Let $\phi : G \rightarrow \bar{G}$ be a homomorphism. If $|\ker\phi | = n $ then $\phi$ is an $n$-to-1 mapping from $G$ onto $\phi(G)$.
Approach:
Since $|\ker\phi | = n$, then $\ker\phi = \{g_1,...,g_n\}$ and $\phi(g_i) = e$ Therefore $\phi^{-1}(e) = g_i\ker\phi$ for all $i = 1,\dots,n$. So all the cosets of $\ker\phi$ have the same number of elements. Is this enough to conclude the result? Or do I need to explain more?