Let $H$ be an infinite dimensional complex (or real) Hilbert space, and let $U(H)$ be the unitary (or orthogonal) group. We equip $U(H)$ with the strong topology.
Now, suppose that $\phi: U(H) \rightarrow U(H)$ is a continuous group automorphism.



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> Is it true that $\phi$ is automatically continuous with respect to the norm topology on $U(H)$?
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I tried proving this directly, but I did not get very far.
Next, I tried characterizing the automorphisms of $U(H)$ more abstractly.
I vaguely recall reading somewhere that any automorphism $\phi$ of $U(H)$ can be realized by conjugation by either an element of $U(H)$ itself, or by conjugation by a conjugate linear isometry of $H$, which would thus automatically imply that $\phi$ is continuous with respect to the norm topology, but I haven't been able to find this result or prove it myself.



On the automorphisms of the unitary group in the strong operator topology
This result is true, and is surprisingly (to me) nontrivial. One reference I could come up with is the paper " Transformations of the unitary group on a Hilbert space" by L. Molnar and P. Semrl. Please look at Theorem 2.5 in this paper : <https://pdfs.semanticscholar.org/386f/31da52d66592ccee13ed046cac4d9ed2444f.pdf>


The authors credit the result to M. Broise "Commutateurs dans le groupe unitaire d’un facteur"(Cor 3).


(I didn't check Broise's paper though, as it's in a language that I can't read :)).


