Let $\mathcal{A}$ be a $C^\*$ algebra. Let $(\pi, \mathcal{H})$ be a faithful, irreducible, unitary, Hilbert space representation of $\mathcal{A}$; i.e., $\pi:\mathcal{A}\rightarrow\mathcal{B}(\mathcal{H})$ is an injective \*-homomorphism from $\mathcal{A}$ into a subset of the bounded operators $\mathcal{B}(\mathcal{H})$ on a Hilbert space $\mathcal{H}$, such that the resulting representation is irreducible. (A representation is irreducible if whenever $\pi(\mathcal{A})$ acts invariantly on a subspace $\mathcal{H}^\prime \subseteq \mathcal{H}$, then $\mathcal{H}^\prime = \mathbf{0}$ or $\mathcal{H}^\prime = \mathcal{H}$.)


Let $U \in \mathcal{B}(\mathcal{H})$ be any unitary ($U^\* = U^{-1}$) bijection on $\mathcal{H}$, not necessarily in $\pi(\mathcal{A})$.


Does it follow that $U\pi(\mathcal{A})U^{\*} = \pi(\mathcal{A})$?



Do unitary bijections act invariantly on irreducible representations?
The answer is no.


Consider the Toeplitz algebra $\mathcal T$ with its canonical representation on $\ell^2 \mathbb N$, which is generated as a $C^\star$-algebra by the shift $S(e\_n)=e\_{n+1}$. It is well-known that the Toeplitz algebra contains all compact operators; hence the representation is irreducible.


It is well-known that any other isometry $V$ of Fredholm index $-1$ is conjugate to $S$. Now, it is easy to see that if $V$ is in $\mathcal T$, then $[S,V]$ is compact. Indeed, the Toeplitz algebra modulo the compact operators is just $C(S^1)$, which is commutative. It is easy to construct such $V$ with $[S,V]$ not compact.


