The rings $\mathbb{Z}/n$ are the only examples.
I assume that "primitive character" just means that it is faithful, or equivalently that it does not factor through a proper quotient; this is the meaning for Dirichlet characters, I think.
The fact that this is true for $R = \mathbb{Z}/n$ is sort of a coincidence. It comes from the fact that
- the primitive characters of $\mathbb{Z}/n$ can canonically be identified with the primitive $n^\text{th}$ roots of unity in $\mathbb{C}$, and there are $\varphi(n)$ of these, and
- the unit group $(\mathbb{Z}/n)^{\times}$ of $\mathbb{Z}/n$ also has size $\varphi(n)$.
These can't be canonically identified (the latter is a group but the former isn't). The unit group does act freely and transitively on the primitive $n^\text{th}$ roots of unity, because it is also the automorphism group $\operatorname{Aut}(\mathbb{Z}/n)$, and the Galois group $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, but to get an identification from here requires choosing a primitive $n^\text{th}$ root of unity, and to identify $\operatorname{Aut}(\mathbb{Z}/n)$ with the unit group involves the isomorphism $\operatorname{End}(\mathbb{Z}/n) \cong \mathbb{Z}/n$, where $\mathbb{Z}/n$ is an abelian group on the LHS and a ring on the RHS. So there are four interesting sets here all of which have cardinality $\varphi(n)$, as well as two different versions of $\mathbb{Z}/n$, the group and the ring.
In general the number of primitive characters of the additive group of $R$ is zero; among the finite abelian groups only the finite cyclic groups can have primitive characters (since these are the only finite subgroups of $\mathbb{C}^{\times}$). It is only nonzero if $(R, +) \cong \mathbb{Z}/n$, meaning it must be additively generated by a single $r \in R$, meaning $1$ must be an integer multiple of $r$, say $1 = kr$. This means $k$ and $r$ must be units $\bmod n$ so $r = k^{-1} \cdot 1$ and $R \cong \mathbb{Z}/n$ as a ring.
