Let $A$ be a (non-unital) $C^\*$-algebra with multiplier $C^\*$-algebra $M(A)$. Let $\phi: M(A) \to M(A)$ be a $\*$-automorphism. Is it true that $\phi$ is automatically strictly continuous (on bounded subsets)?


Some remarks/observations:


(1) If $A = B\_0(H)$, then this is true because $\*$-automorphisms of $B(H) = M(B\_0(H))$ are automatically strict (as they are given by conjugation with a unitary).


(2) If $A$ is separable, this is true due to a result by Woronowicz which says that
$$A= \{x \in M(A): xM(A) \mathrm{\ is \ separable}\}$$
so that we can reconstruct $A$ from its multiplier $C^\*$-algebra $A$.


(3) I tried to see what happens in the commutative case, so $A=C\_0(X)$. Then a $\*$-automorphism of $M(A) = C\_b(X)= C(\beta X)$ corresponds to a homeomorphism $\beta X \to \beta X$. I have hope that if a counterexample exists, then a smart example of such a homeomorphism can lead to a counterexample.


(4) The following question seems to be related: Does a strict $\*$-automorphism $\phi: M(A) \to M(A)$ preserve the subalgebra $A$, i.e. do we have $\phi(A)\subseteq A?$



Is a $*$-automorphism $M(A) \to M(A)$ automatically strictly continuous?
I think that the answer is no.


Let $\mu$ be a non-trivial homeomorphism of $\beta \bf N$ with distinct points $y,z\in \beta\bf N\setminus \bf N$ such that $\mu(y)=z$ and $\mu(z)=y$. Set $A=\{f\in C(\beta{\bf N} ): f(y)=0\}$. Then $M(A)=C(\beta {\bf N})$. Let $\phi: M(A)\to M(A)$ be given by $\phi(f)(x)=f(\mu(x))$ $(f\in C(\beta {\bf N}), x\in \beta {\bf N})$.


Let $(f\_{\alpha})$ be a bounded approximate identity for $A$. Then $(f\_{\alpha})$ converges strictly to $1\in M(A)$ but $(\phi(f\_{\alpha}))$ does not converge strictly to $\phi(1)=1$. To see this, take $g\in A$ such that $g(z)=1$. Then $\phi(f\_{\alpha})(z)g(z)=\phi(f\_{\alpha})(z)=f\_{\alpha}(\mu(z))=f\_{\alpha}(y)=0$ but $\phi(1)(z)g(z)=1$. Hence $\phi(f\_{\alpha})g$ does not converge in norm to $g$.


