What is the group of automorphisms of $l^{\infty}$? What is the group of automorphisms of $l^{\infty}$?
I think it would be the permutations of the integers. Is this right?
 A: Yes, the group of automorphisms of $\ell^\infty(\mathbb{Z})$ preserving the $W^*$-algebra structure is the group of permutations of $\mathbb{Z}$.
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 - that is, homomorphisms $\chi: \ell^\infty(\mathbb{Z}) \to \mathbb{C}$ - to characters, so it defines a permutation of the characters.  By the Gelfand-Naimark correspondence, there is a compact Hausdorff space $\beta\mathbb{Z}$ called the Stone-Cech compactification of $\mathbb{Z}$ such that 
$$\ell^\infty(\mathbb{Z}) \cong C(\beta\mathbb{Z}) $$
Points of $\beta\mathbb{Z}$ are the same as characters  $\chi: \ell^\infty(\mathbb{Z}) \to \mathbb{C}$.  Every point $n \in \mathbb{Z}$ defines a character $\chi$ by
$$ \chi(f) = f(n) $$
Not all characters are of this form, as pointed out by Yemon Choi and Uri Bader.  However, the weak-$\ast$-continuous characters are.   If $\alpha$ is a $W^*$-algebra automorphism of $\ell^\infty(\mathbb{Z})$ and $\chi$ is a weak-$\ast$-continuous character, $\chi \circ \alpha$ is again weak-$\ast$-continuous.  So, every $W^*$-algebra automorphism of $\ell^\infty(\mathbb{Z})$ defines a permutation of the integers.
The concept of 'weak-$\ast$-continuous character' is a bit fancy, so I like 
Nik Weaver's simpler argument below.  Just as the integers give the weak-$\ast$-continuous characters on $\ell^\infty(\mathbb{Z})$, they give the isolated points in $\beta\mathbb{Z}$ - but this is easier to understand, and easier to prove.  By the functorality of the Gelfand-Naimark correspondence any automorphism of $\ell^\infty(\mathbb{Z})$ comes from a homeomorphism of $\beta\mathbb{Z}$, and these clearly map isolated points to isolated points.  So, any automorphism of $\ell^\infty(\mathbb{Z})$ gives a permutation of the integers.
A: This is to record what is probably the easiest proof (that the answer is "yes"), following a remark by Nik Weaver.
Noting that $\ell^\infty(\mathbb{Z})\simeq C(\beta\mathbb{Z})$, using Banach-Stone, we see that every auto gives a homeo of $\beta\mathbb{Z}$.
But since $\mathbb{Z}$ is the subset of isolated points, every homeo preseves it, and since it is dense the homeo is determined by its restriction to $\mathbb{Z}$. 
A: For a standard Lebesgue space $(X,\mu)$ the autmorphism group of the W* algebra $L^\infty(X,\mu)$ is the group of measure class preserving automorphisms of $(X,\mu)$ - the group of a.e defined measurable map $\sigma:X\to X$ s.t $\sigma_*\mu$ and $\mu$ are equivalent, modulo the relation of being identical a.e.
(more generally: the category of standard Lebesgue spaces and (the op of) the category of commutative W* algebras with separable predual are equivalent).
Apply this general fact in the case $(\mathbb{Z},\text{counting})$ and get the answer: yes, every automorphism is given by a permutation.
