Finite Modules Isomorphism Two vector spaces are isomorphic if and only if they have the same dimension.
In particular, two vector spaces over a finite field are isomorphic if and only if they have the same cardinality.
For a general ring $R$, I think it is not true that two $R$-modules are isomorphic if they have the same cardinality.
I would like to have some example of that with a finite ring $R$, so I can have a better picture of the situation.

Thus, I am looking for examples of the following.
  
  
*
  
*A finite ring $R$ and two $R$-modules $M$ and $N$ of the same finite cardinality such that $M \ncong N$.
  
*A similar example where $M$ and $N$ are moreover isomorphic as abelian groups.

 A: Any abelian group with exponent dividing a positive integer $n$ is a $\mathbb Z/n \mathbb Z$ module. This, together with the structure theorem for finite abelian groups, givens plenty of examples. For instance,
$$
\mathbb Z/n\mathbb Z \oplus \mathbb Z/n\mathbb Z \ncong \mathbb Z/n^2 \mathbb Z
$$
as $\mathbb Z/n^2\mathbb Z$-modules, but the two modules have the same cardinality.
Let us construct two modules over a finite ring $R$ having isomorphic group ($\mathbb Z$-module) structure but non-isomorphic $R$-module structure.
Let $R$ be a ring (not necessarily finite at this point) and $S=R[X]/(X^2)$. 
The $R$-module $R^2$ is an $S$-module in two different ways: First $S\cong_R R^2$ gives a structure where $X(1,0)=(0,1)$ and $X(0,1)=(0,0)$. Second $S/(X)\cong R[X]/(X)$ induces an $S$-module structure on $R$. Taking a direct sum, we obtain another $S$-module structure on $R^2$ where now $X(1,0)=(0,0)$ and $X(0,1)=(0,0)$. The two $S$-module structures are different, as $X$ acts differently on them. However, the $R$-module structure is the same.
In particular, therefore also the module structure over the prime ring of $R$ is the same. However, this is simply the group structure. Taking $R$ a finite ring produces the desired example.
