On the form of ring homomorphism from $Z_{n}$ to itself.

$\phi$ : $Z_{n} \rightarrow Z_{n}$, $x \mapsto ax$, where $a^{2} = a$ is ring homomorphism.

Let $\phi(1) = a$.
Then $a = \phi(1) = \phi(1)\phi(1) = a^{2}$.

Let $x \in Z_{n}$.
Then $\phi(x) = \phi(1 + \dots + 1) = \phi(1)x = ax$.

Hence, our original conjecture is true.