When giving $A_1,A_2$ two isomorphic maximal commutative semi-simple sub algebras of $M_n(\mathbb{C})$, are these algebras conjugate in $M_n(\mathbb{C})$? Namely, does there exists a regular matrix $P$ such that $P^{-1}A_1P=A_2$?

The answer is yes. Each corresponds to a decomposition of the identity as a sum of mutually orthogonal primitive idempotents, each of which has rank 1.
– Geoff RobinsonApr 14 '13 at 11:21

Thank you for that answer. Where did you use the isomorphism condition? Are all the semi-simple commutative maximal sub-algebras of M_n(C) isomorphic? If I understand your answer, the only maximal commutative semi-simple sub-algebra of M_(C) is the diagonal algebra?
– ofirApr 14 '13 at 12:09

@ofir: Any such $A$ is a product of copies of $\mathbf{C}$, so a faithful representation $A \hookrightarrow {\rm{M}}_n(\mathbf{C})$ on $\mathbf{C}^n$ from a commutative semisimple $\mathbf{C}$-algebra $A$ sends the primitive idempotents to {\em distinct} pairwise orthogonal commuting nonzero idempotent linear operators on $\mathbf{C}^n$ whose sum is the identity operator. This is exactly a decomposition of $\mathbf{C}^n$ as a direct sum of nonzero subspaces. The "maximal" way to do this is with an ordered $n$-tuple of independent lines, and in a suitable basis all $n$-tuples look the same...
– user30379Apr 14 '13 at 14:47

Thank you for your answers. Are the same arguments hold when replacing M_n(C) with twisted group algebra. Or maybe even for any semi-simple algebra? That is When giving A_1,A_2 two Isomorphic maximal commutative semi-simple sub algebras of a twisted group algebra A (or any semi-simple algebra). Are these algebras conjugate in A.
– user33117Apr 15 '13 at 6:31