# *-automorphisms of von Neumann algebras

It should be a very simple question, but I couldn't find any reference concerning it.

Let $$\mathcal{H}$$ be a separable Hilbert space, $$\mathcal{A}\subset\mathcal{B}(\mathcal{H})$$ a von Neumann algebra and $$\phi:\mathcal{A}\rightarrow\mathcal{A}$$ an *-automorphism (linear map, preserving products and adjoints). The question is: does it exist a unitary operator $$U\in\mathcal{B}(\mathcal{H})$$ such $$\phi(A)=UAU^*$$ for all $$A\in\mathcal{A}$$.

I'm looking for a proof, a reference or a counterexample about the above question.

A remark about continuity: If I'm not confused, every *-automorphism (*-homomorphism indeed) between vN algebras is always continuous in the operator topology, but it may not be necessary continuous in the $$\sigma$$-weak topology. If the above statement is false in general, may it be true for $$\sigma$$-weak continuous *-automorphisms?

If you had that property, it would mean that every automorphism of every von Neumann algebra extends to an automorphism of $$B(H)$$. That's very non true. For instance, let $$p$$ be a finite-projection, and consider the algebra $$A=\{\alpha p+\beta(1-p):\ \alpha,\beta\in\mathbb C^2\}.$$ This algebra is of course $$\mathbb C^2$$ in disguise, and it has the automorphism $$(\alpha,\beta)\longmapsto (\beta,\alpha)$$. If you had your unitary, you would have $$UpU^*=1-p$$, impossible since $$p$$ is finite and $$1-p$$ is infinite. The automorphism is continuous in any reasonable topology.
Note that the example above even works in finite dimension, just by choosing the projection $$p\in M_n(\mathbb C )$$ with $$\operatorname{Tr}(p)\ne n/2$$.