Let $\phi$ define a $\*$-automorphism from the matrix algebras $M\_n(\mathbb{C})$ to $M\_n(\mathbb{C})$ such that $\phi(I) = I$. Is it true that any such map $\phi$ can be represented as $\phi(x) = U x U^{\dagger}$ (where $U$ is a suitable unitary matrix)? If not, what is the most general expression?



Representation of $*$-automorphism on finite dimensional matrix algebras
Here is one generalization:



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> Every $\*$-automorphism of the algebra of compact operators on a Hilbert space is conjugation by a unitary operator on that space.
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Using the fact that the algebra of compact operators is irreducible, this can be seen as a special case of:



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> Every irreducible $\*$-representation of the algebra of compact operators on a Hilbert space is unitarily equivalent to the identity representation.
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A proof can be found for instance in Section 1.4 of Arveson's *[An invitation to C`*` algebras](http://books.google.com/books?id=zRep59fhoGkC&lpg=PP1&dq=arveson%2520invitation&pg=PA20#v=onepage&q&f=false)*. Another proof of the first assertion that gives more information can be found in Proposition 1.6 of Raeburn and Williams's *[Morita equivalence and continuous trace C`*`-algebras](http://books.google.com/books?id=KV_cC5uYA7EC&lpg=PP1&pg=PA5#v=onepage&q&f=false)*.


The first part is still true if you take all bounded operators instead of only the compact ones. (And these are the same thing in the finite dimensional case.)


