Theorem   Two cyclic rings are isomorphic if and only if they have the same order and the same behavior.

Proof. Let $R$ be a cyclic ring with behavior $k$ and $r$ be a generator of the additive group of $R$ with $r^2=kr$ Also, let $S$ be a cyclic ring.

If $R$ and $S$ have the same order and the same behavior, then let $s$ be a generator of the additive group of $S$ with $s^2=ks$ Define $\varphi \colon R \to S$ by $\varphi(cr)=cs$ for every $c \in \mathbb{Z}$ This map is clearly well defined and surjective. Since $R$ and $S$ have the same order, $\varphi$ is injective. Since, for every $a,b \in \mathbb{Z}$ $\varphi(ar)+\varphi(br)=as+bs=(a+b)s=\varphi((a+b)r)=\varphi(ar+br)$ and

it follows that $\varphi$ is an isomorphism.

Conversely, let $\psi \colon R \to S$ be an isomorphism. Then $R$ and $S$ must have the same order. If $R$ is infinite, then $S$ is infinite, and $k$ is a nonnegative integer. If $R$ is finite, then $k$ divides $|R|$ which equals $|S|$ In either case, $k$ is a candidate for the behavior of $S$ Since $r$ is a generator of the additive group of $R$ and $\psi$ is an isomorphism, $\psi(r)$ is a generator of the additive group of $S$ Since $(\psi(r))^2=\psi(r^2)=\psi(kr)=k\psi(r)$ it follows that $S$ has behavior $k$