Do the fusion categories $Rep(S_4)$ and $Rep(A_5)$ admit nonsymmetric braidings? All the other rep. cats. of finite subgroups of $SU(2)$ do (in the McKay correspondence). My guess is no.
Eric,
The answer is no for $Rep(A_5)$ and yes for $Rep(S_4)$, thanks to Victor Ostrik's observation. For a braided category $C$ let $C'$ denote its Mueger center, i.e., the subcategory of objects $Y$ in $C$ such that the square of braiding of $Y$ with any $X$ in $C$ is identity. So $C$ is symmetric if $C=C'$ and $C$ is nondegenerate (or modular) if $C'$ is trivial.
Note that $C:=Rep(A_5)$ is simple, i.e., it has no nontrivial proper fusion subcategories. Now if $C$ has a nonsymmetric braiding then $C' \neq C$ is a proper subcategory. So $C'$ is trivial, i.e., $C$ with the above braiding is nondegenerate (modular). This cannot happen (e.g., $C$ has a simple object of dimension 5, but in a modular category the square of dimension of any object divides dimension of the category, thanks to the result of EtingofGelaki).
For $D:= Rep(S_4)$ there is a nonsymmetric braiding with $D'=Rep(S_3)$, namely the equivariantization of a pointed category $Vec_{Z/2Z\oplus Z/2Z}$ with respect to an action of $S_3$.

1$\begingroup$ Thanks Dimitri! How do you rule out $D^\prime=Rep(S_3)$? $\endgroup$ – Eric Rowell May 28 '10 at 16:33

3$\begingroup$ Dmitri, for $D=Rep(S_4)$ there is one more possibility, namely $D'=Rep(S_3)$. Actually I believe that for $D=Rep(S_4)$ the alternative braiding is possible. Namely consider pointed category with underlying group $Z/2Z \oplus Z/2Z$ and quadratic form which takes value 1 on a nontrivial element. Then equivariantize with respect to the obvious $S_3$action. $\endgroup$ – Victor Ostrik May 28 '10 at 16:40

$\begingroup$ Victor is right. I will edit the above answer. $\endgroup$ – Dmitri Nikshych May 28 '10 at 16:49
You can also answer the question using the classification of Rmatrices over group algebras. Indeed, the Rmatrices in the Hopf algebra $ kG $, are classified by pairs $ (A, \rho) $ where $ A \subset G $ is an abelian normal subgroup and $ \rho: A \times A \to k^* $ is a bilinear form ad $ G $invariant. It is easy to see that the associated Rmatrix is symmetric if and only if the bilinear form is skewsymmetric.
It is now clear that $ kA_5 $ does not have Rmatrices in addition to the trivial. On the other hand, the only normal abelian subgroup of $ S_4 $ is formed by the union of the conjugacy classes of $(12)(34)$ and the identity of $S_4$. This normal subgroup is isomorphic to $C_2 \times C_2 $. As was mentioned by Ostrik, one example could be the next: the quadratic form that takes values 1 on nontrivial elements. That bilinear form satisfies the conditions and defines a nonsymmetric Rmatrix.

$\begingroup$ Thanks Cesar. What is a reference for this classification? Is there a variation for twisted group algebras? So this might lead to a more general statement for categories with the same fusion rules as $Rep(G)$ for $G$ a nonabelian simple group? $\endgroup$ – Eric Rowell Jun 3 '10 at 13:30

$\begingroup$ 1) You can see the result in Example 2.1 of arXiv: math/0605731, or arXiv: qalg/9706007. 2) On the variation for the twisted group algebras, I do not know anything, but I do not know which would be a possible variation. 3) The result implies that the only braid on Rep (G) for G a nonabelian simple group is the trivial braid. But I think you can not say anything specific for fusion categories with the same fusion rules of Rep(G), since there are pairs of fusion categories with the same rules of fusion, where one of them is braided and the other has not a braid. $\endgroup$ – César Galindo Jun 3 '10 at 15:03

$\begingroup$ Cesar, I was thinking of using Prop. 9.11 of 0809.3031 to say something about when the only braidings on a category are symmetric. Thanks for the references. $\endgroup$ – Eric Rowell Jun 3 '10 at 18:30