# Commutative subalgebras of $B(H)$ whose all automorphisms are in the form of unitary conjugation

Let $$H$$ be a complex Hilbert space.

Is there a compact Hausdorff space $$X$$ such that $$C(X)$$ is embeded in $$B(H)$$ and for every homeomorphism $$\alpha$$ of $$X$$ there exist a unitary operator $$u\in B(H)$$ with $$u^*fu=f\circ \alpha^{-1}\quad \forall f \in C(X)\subset B(H)$$?

The motivation comes from the following post:

Can every dynamical system be interpreted in terms of (unitary) conjugation in an operator algebra

• Yes, the one point space. I assume you actually wanted to ask something more specific? – Yemon Choi Aug 14 at 1:41
• @YemonChoi Yes I mean non degenerate space. – Ali Taghavi Aug 14 at 1:50
• @YemonChoi a space with at least 3 point. However infinite space or even connected space would be ideal examples. – Ali Taghavi Aug 14 at 1:54
• @YemonChoi One can generalize the question in a more algebraic manner as follows: For a goven $C^*$ algebra $A$, one can think to all subalgebras $B$ such that every automorphism of $B$ is in the form of a unitary conjugation. – Ali Taghavi Aug 14 at 2:50
• Can't you just take any faithful representation of the $C^*$-algebra $C(X) \rtimes {\rm Homeo}(X)$? – Jesse Peterson Aug 14 at 15:35