Do character tables determine association schemes up to isomorphism? I am interested in commutative, but not-necessarily-symmetric association schemes. I noticed that the conjugacy class scheme of the dihedral group of order 8 has the same character table as the one for the quaternion group, and the schemes are also isomorphic.
Are there examples of non-isomorphic association schemes that have the same character table? I am not married to association schemes arising from groups.
 A: IMHO the isomorphism of association schemes is understood as isomorphism of underlying coloured digraphs. The smallest example of non-isomorphic (in this sense) association schemes with the same character table comes from a pair of nonisomorphic degree 6 strongly regular graphs on 16 vertices. One of these graphs is Shrikhande graph, the other is $4\times 4$ rook graph. 
In the example in the question one has association schemes of conjugacy classes of groups $D_8$ and $Q_8$, and they happen to be isomorphic in the sense of coloured digraph isomorphism. It is example number 9 in tables by Izumi Miyamoto and Akihide Hanaki.
In fact if isomorphism of association schemes was possible to decide from their character tables then the graph isomorphism problem would be easy to solve.
A: If I understand right the character table of an association scheme is the table of irreducible representations of the Bose-Mesner algebra.
If we can recover this algebra, and the adjaceny matrices within it, we can recover the association scheme using the formula $A_i A_j = \sum_k p_{ij}^k A_k$, which uniquely determines $p_{ij}^k$ as the $A_k$ form ab asis.
As long as this algebra is commutative and semisimple, then one can recover this algebra and those elements from the character table: The algebra is just the product of one copy of $\mathbb C$ for each character, and the class of an adjacency matrix is just the ordered tuple of the values of the different irreducible characters.
Since your association scheme is commutative, the algebra is indeed commutative and also from looking it up appears to be semisimple.
Then it's easy to get the identity, as $0$ is the only $i$ such that $p_{ij}^k=0$ for all $j \neq k$, and then get $i^T$ using the fact that $i^T$ is the only $j$ such that $p_{ij}^0\neq 0$.
