Yes, the group of automorphisms of $\ell^\infty(\mathbb{Z})$ preserving the $W^*$-algebra structure is the group of permutations of $\mathbb{Z}$.
It's pretty easy to see this. On the one hand, any permutation of $\mathbb{Z}$ acts as an automorphism of $\ell^\infty(\mathbb{Z})$. On the other hand, any $W^*$-algebra automorphism of $\ell^\infty(\mathbb{Z})$ sends characters (homomorphisms to $\mathbb{C}$) to characters, so it defines a permutation of the characters. But the characters of $\ell^\infty(\mathbb{Z})$ are all of the form $f \mapsto f(n)$ for some integer $n$. (This follows from the Gelfand-Naimark theorem and a little work.) So, any $W^*$-algebra automorphism of $\ell^\infty(\mathbb{Z})$ comes from a permutation of the integers.
We could say $C^*$-algebra instead of $W^*$-algebra in the last paragraph and everything would still be true.