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 $\phi(A)\subseteq A?$