Find all homomorphisms from $\mathbb Z_{12}$, the cyclic group of order $12$, to $\mathbb Z_6$. For each homomorphism $f\colon \mathbb Z_{12}\to \mathbb Z_6$, determine the kernel $\ker(f)$ and the image $f(\mathbb Z_{12})$.
I've already used the fact that the generators of $\mathbb Z_{12}$ are $\bar 1, \bar 5, \bar 7, \bar 11$ and the generators of $\mathbb Z_6$ are $\bar 1$ and $\bar 5$. To get the homomorphisms which lead to the $\mathbb Z_6$ outputs of $(\bar 0, \bar 5, \bar 4, \bar 3, \bar 2, \bar 1, \bar 0, \bar 5, \bar 4, \bar 3, \bar 2, \bar 1)$ and $(\bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5, \bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5)$ but I don't know where to go from here to obtain the rest.
A homomorphism $\mathbb Z_{12}\to \mathbb Z_6$ is completely determined by $f(\bar 1)$ as then necessarily $f(\bar n)=f(n\cdot \bar 1)=n\cdot f(\bar 1)$. All you need to ensure is that $12\cdot f(\bar 1)=0$, but that is clear as it holds for any element of $\mathbb Z_6$.