If $ f: \mathbb Z_{m} \to \mathbb Z_{n} $ is a ring homomorphism, then $f(0) = 0, mf(1) = 0, f(1)^2 = f(1)$. This is easy to show.
But, is the converse true?
Edit-1 :
In the below comments it is shown that the answer is a clear no. i.e, the converse does not hold. But, my doubt is, what is the justification for the answer given by user26857 in the following post to use the above identities to find the number of homomorphisms from $\mathbb Z_{m}$ to $ \mathbb Z_{n}$: The number of ring homomorphisms from $\mathbb{Z}_m$ to $\mathbb{Z}_n$ ?
Expanding a bit on the comment of Badam Baplan:
Let $f:\mathbb{Z}_m\rightarrow \mathbb{Z}_n$ be a group homomorphism. Since $1$ generates the group $\mathbb{Z}_m$, the map $f$ is completely determined by $f(1)$. Now notice that $mf(1)=0$, hence the order $\text{ord}(f(1))$ of $f(1)$ divides $m$. On the other hand $nf(1)=0$ since $n$ is the order of $\mathbb{Z}_n$, it follows that the order $\text{ord}(f(1))$ divides $n$ as well. It follows that $\text{ord}(f(1))\mid \gcd(m,n)$.
Hence if $m,n$ are coprime, it follows that $\text{ord}(f(1))=1$ which in turn yields $f(1)=0$. It follows that $f=0$. Hence in general there are not to many group morphisms between $\mathbb{Z}_m$ and $\mathbb{Z}_n$, let alone ring morphisms.