Let $(R,+,\cdot)$ be a finite ring. Obviously, $(R,+)$ is an abelian group; however the unit (multiplicative) group need not be abelian.
My question is the following problem:
Given the dual group $\widehat{R}$ of $(R,+)$, under what conditions of $R$ is it true that
$$ \#\{\text{primitive* characters in ($\widehat{R},\cdot$)} =\#R^\times, $$ i.e. when does the number of primitive* characters of the abelian group $(R,+)$ equal the number of units in $R$?
*((DefinitionAdded Given the dual group $\widehat{R}$ of $(R,+)$, character)). A character $\chi\in\widehat{R}$ is is said to be primitive with respect to a collection primitive with respect to a collection $\mathcal{C}$ of subgroups$\mathcal{C}$ of subgroups of $(R,+)$,$(R, +)$ if $\chi$ does not descend to a character on any quotient group $R/S$ for all $S\in \mathcal{C}$$S\in\mathcal{C}$. To descend here is canonical in the sense that $\chi$ is invariant on the elements of any coset of $R/S$.
My question is the following problem:
Are there any generic non-trivial conditions on $\mathcal{C}$ such that $$\#\{\text{primitive* characters in $\widehat{R}$ w.r.t $\mathcal{C}$}\}=\#R^\times,$$ i.e. when does the number of primitive* characters of the abelian group $(R,+)$ w.r.t $\mathcal{C}$ equal the number of units in $R$?
NOTES
The archetypal example is the ring $\mathbb{Z}/n\mathbb{Z}=(\{0,1,2,\dotsc,n-1\},+,\cdot)$ of integers modulo $n$ where the answer is positive when one considers all proper quotients.
(Added later) Qiaochu Yuan's answer shows that $\mathbb{Z}/n\mathbb{Z}$ is indeed the only example when one considers$\mathcal{C}$ contains all proper subquotients. My question is whether a well-defined non-trivial collection of, rather than$\mathbb{Z}/d\mathbb{Z}$ for all, quotients leads to an affirmation of the equality above $d$ divides $n$.