Under what conditions two cyclic modules are isomorphic?

(I assume $R$ is associative, guessing it's implicit.) Write $L_c(r)=cr$: then $L_c$ is an endomorphism of $R$ as right $R$-module, and $L_{cd}=L_c\circ L_d$ for all $c,d$.

Equivalences:

(a) $R/I$ and $R/J$ are isomorphic right $R$-modules.

(b) there exists $a\in R$ such that $L_a^{-1}(J)=I$ and $aR+J=R$.

(c) there exist $a,b\in R$ such that
(c1) $aI\subset J$
(c2) $bJ\subset I$
(c3) $ba-1\in I$
(c4) $ab-1\in J$.

If (a) holds, consider an isomorphism $q:R/I\to R/J$, and lift the image of $1$ as an element $a\in R$. Then $q$ is induced by $L_a$ and (b) follows. Also, choose a lift $b$ of the image of $1$ by the $q^{-1}:R/J\to R/I$. Then (c) holds.

Suppose that $a,b$ as in (c) exist. Then by (c1) $L_a$ induces an homomorphism $q:R/I\to R_J$, by (c2) $L_b$ induces an homomorphism $q':R/J\to R_I$. So $L_{ba}$ and $L_{ab}$ induce endomorphisms of $R/I$ and $R/J$, which by (c3) and (c4) are the identity. Hence $q$ and $q'$ are inverse to each other. So (a) holds.

Similarly if $a$ exists as in (b) then by the first half of (b), $L_a$ induces an injective homomorphism $R/I\to R/J$, which is surjective by the last part.