Is there a 2 component link with full symmetry? If you take a (labelled, oriented) 2 component link, it has a symmetry group which is a subgroup of the 16 element group $Z2 \times (Z2 \times Z2 \ltimes S_2)$ (mirror, reverse each component, swap the components). With Parsley, Cornish, and Mastin (http://arxiv.org/abs/1201.2722) I compiled a table of the frequency with which each conjugacy class of subgroups of this group comes up as a symmetry group for links up to 14 crossings.
Interestingly, the full group never came up as a symmetry group! Can a 2 component link ever have the full 16 element group? Is there some obstruction to this? Can anyone propose an example link?
Edit: A non split example. It is easy to construct split ones!
Clarification: The traditional symmetry group is the mapping class group $MCG(S^3,L)$. The symmetry group I want is the image of that group under the homomorphism $MCG(S^3,L) \rightarrow MCG(S^3) \times MCG(L)$, which is the 16 element group above. So another way to ask the question would be: can this homomorphism ever be surjective?
 A: This is more a comment on Ryan's nice answer than an independent answer itself, but it's too long for a comment box.
Let $K$ be an oriented knot in the solid torus which is isotopic (in the solid torus) to its reverse.  [Edit: It should also be amphichiral, as well as invariant under the map of the solid torus which reverses the $S^1$ factor.]  Let $L$ be a two component unoriented framed link in $S^3$ which is isotopic to its mirror image (without exchanging the two components) and also has an isotopy which exchanges the two components.  Then replacing each component of $L$ with a copy of $K$ yields a two component oriented link in $S^3$ which has the full order 16 symmetry mentioned in the question.
In Ryan's example, $K$ is what I would call a Whitehead double with a funky figure-8 clasp, and $L$ is the Hopf link.
It's relatively easy to think of $K$ and $L$ satisfying the above conditions, so one can construct a large family of examples this way.
A: I think there is a non-hyperbolic link that does the job. 
The link that I'm thinking of could be called the splice of two Bing doubles of a figure-8 knot.  Another way to describe this link is to start with this 4-component link:  and splice two figure-8 knots against it.  You'd splice the knots against the red and purple components, respectively. 
above image taken from Morwen Thistlethwaite's homepage.  Below is a picture of the link I'm referring to:

The JSJ-decomposition of the complement of this link has 3 incompressible tori, separating the link complement into two Borromean ring complements, and two figure-8 knot complements.  There is an orientation-reversing diffeomorphism of $S^3$ that preserves the link, since the Borromean rings admit one, and the figure-8 complements do, too.
As you can see from the way I've drawn it, there is a symmetry that permutes the two components, preserving the orientation of $S^3$, and from the Thistlethwaite diagram you can see a symmetry that reverses the orientation of either component, also preserving the orientation of $S^3$ -- this needs that the figure-8 knot is invertible. 
