Uniqueness of the fusion ring for simple finite group We know that the irreducible representations $R_i$ of a group $G$ can give rise to a fusion ring: $R_i\otimes R_j = \oplus_k N^{ij}_k R_k$.
I wonder if the following statement is true or not:
If $G$ is a simple finite group, then its fusion ring is different from the fusion rings of any other groups.
 A: The fusion ring, as a ring with basis, contains the same information as the character table. So your question, phrased in language more familiar to finite group theorists, is:

Is a finite simple group determined by its character table among all finite groups? 

Here is how to see the above claim. The easy direction is that if you know the character table then you know the fusion ring, as a ring with basis: this is clear since multiplication of characters corresponds to tensor product of representations. In the other direction, for varying $i$, the matrices $N^{ij}_k$ admit simultaneous eigenvectors (namely the vectors $g \mapsto \chi_V(g)$, where $V$ runs over all irreps), and the simultaneous eigenvalues for these eigenvectors are the character values of $R_i$. 
It's known that a finite simple group is determined by its character table among all finite simple groups. And the character table determines whether or not a finite group is simple: $G$ is simple iff the kernel of every nontrivial irreducible character (the elements $g$ such that $\chi(g) = \chi(1)$) is trivial, since kernels are always normal subgroups and every proper normal subgroup is contained in a kernel. 
So the final answer is yes. 
A: If I've understood your question correctly, you are asking whether two simple groups can have unitally order isomorphic representation rings (where the cone for the partial ordering is given by the actual characters, generated additively by the irreducible characters). There is a partial result available, and group theorists can probably tell us if this is sufficient.
If we view the representation ring of the finite group $G$, $R(G)$, as a partially ordered ring with $1$ (with positive cone generated additively by the irreducible characters), then we can recover the cardinality of $G$ from it. [That is, if $R(G) $ and $R(G_1)$ are isomorphic as unital partially ordered rings, then $G$ and $G_1$ have equal cardinality; for trivial reasons (the rank, as an abelian group), they also have the same number of conjugacy classes, and the same number of irreducible characters.]
Either by the Perron-Frobenious theorem or otherwise, there is a unique positive ring homomorphism $t:R(G) \to {\bf R}$ sending the trivial character to $1$, and this is simply evaluation of the virtual character at $1$, the dimension. Again by the Perron theorem, up to scalar multiple, there is a unique common eigenvector (viewing a virtual character as an endomorphism of $R(G)$ by multiplication) for $t$, specifically, a positive real multiple of the regular representation character, call it $\chi$. In particular, $\chi$ is characterized by being the unique character of $G$ that is a common eigenvector, belongs to $R(G)^+$, and has minimal degree (valuation at $t$) among those satisfying the first two properties. 
It follows that $t(\chi)$ is an invariant of $R(G)$, and of course, $t(\chi) = \chi(1) = |G|$.
With a little more work, the set with multiplicities consisting of the degrees of irreducibles is also an invariant of $R(G)$. [Irreducibles have to be mapped to irreducibles under an  order isomorphism, and evaluate at $t$.]
So if $R(G)$ and $R(G_1)$ are isomorphic as unital partially ordered rings, then $|G| = |G_1|$, they have the same number of irreducibles, and the degrees of their irreducibles are equal (as multisets); of course, the last also implies cardinalities are equal. For simple groups, I don’t know whether this is enough to distinguish them.
As is well known (and probably what motivated the restriction to simple groups), the two interesting groups of order 8 ($D_4$ and the quaternion group) cannot be distinguished by their representation rings.
