Let $R$ be a finite ring and $F$ be an algebraically closed field in which $|R|$ is invertible. Does there exists an $F$-valued character $\chi$ of $(R, +)$ such that every character $\psi$ is of the form $\psi(a) = \chi(ab)$ for some $b \in R$? If not, does the statement hold when $R$ is commutative (or under any other reasonably nice assumption)?

This is true when $R$ is a finite field. I believe that the statement in this case follows from the non-degeneracy of the trace form over the prime field contained in $R$.

Any references would be really helpful.

Motivation: This appears when we study representation theory of unipotent groups. Suppose $U$ be the subgroup of unipotent lower triangular matrices in $GL_n(R)$. Then elements of the form $E_{\chi} = \sum_{M \in U} \chi(M_{2,1})M \in F[U]$ appear naturally where $\chi$ is an $F$-valued character of $(R, +)$. There is nothing special about $M_{2,1}$, one can work with any closed "root subgroup". Now conjugating $E_{\chi}$ by an element of the cartan subgroup yields $E_{\chi(.b)}$ for some $b$ and so it becomes useful to know whether all characters can be obtained in this way.