For fields it is well known, that all fields with finite cardinality $n$ are isomorphic. Does a similar result also hold for rings, i.e. are rings that have the same finite cardinality isomorphic ?
EDIT: Sorry, user Keenan Kidwell got it right, I mean "cardinality", not characteristic.
On
This is not true for fields. $\mathbb{F}_{p^n}$ is a field (in particular, a ring) of characteristic $p$, and so is $\mathbb{F}_{p^m}$, but these fields are isomorphic if and only if $n = m$.
On
Surely not. For example, $\mathbb Z_2$ and $\mathbb Z_2\oplus\mathbb Z_2$ have both characteristic $2$. They cannot be isomorphic, as they have different cardinality.
On
I guess that you're mixing up characteristic with cardinality. Any two fields of the same finite cardinality are isomorphic. On the other hand, the characteristic of a field, defined as the smallest positive integer $n$ such that $1+\cdots+1=0$ ($1$ added to itself $n$ times) in the field, or else if no such integer exists, as $0$, is either $0$ or $p$ with $p$ a prime, so always "finite." But there is no relationship between characteristic and cardinality beyond the fact that a field with finite cardinality must have positive characteristic.
A more dramatic example would be $\mathbb{F}_p$ and $\mathbb{F}_p(t)$ where $t$ is an indeterminate, and $p$ is prime. They both have characteristic $p$, but one of them is finite while the other is not. You can also take an algebraic closure of $\mathbb{F}_p$, instead of $\mathbb{F}_p(t)$.
Later edit: As a counterexaple to your revised statement take $\mathbb{Z}/8\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. They both have the same cardinality, but cannot be isomorphic. One has characteristic $4$, while the other has characteristic $8$.