If you're interested in the involution only defined on the complement, Igor's answer does a fine job. 

But the involution extends to an involution of $S^3$ and perhaps you'd like to see that? 

I think the symmetry is a little tricky to traditionally visualize, because as a map of $S^3$, thought of as the unit sphere $S^3 \subset \mathbb C^2$ it's of the form $(z_1,z_2) \longmapsto (\overline{z_1}, -z_2)$.  If you stereographically project at one of these fixed points, this becomes the involution of $\mathbb R^3$ given by $(x,y,z) \longmapsto (-x,-y,-z)$.  Both the fixed points have to be on the knot, so this means we have to find a "long" embedding of the figure-8 knot in $\mathbb R^3$ which is invariant under the antipodal map.   That's easy.   Here are two such (approximate) positions:


![alt text][1]

![alt text][2]


  [1]: http://etc.usf.edu/clipart/8100/8102/eight_knot_8102_lg.gif
  [2]: http://rybu.org/math/c4.long.2.21.jpg

In the latter picture the fixed point is easier to see -- the blue straight line intersects the knot in 5 points, one at the "center" and four other points indicated by blue balls. 

Looking through the code I used to generate the latter picture, the parametrization I use of the figure-8 knot is:

$$t \longmapsto (5(t^3-3t), 0.25(t^7-42t), t^5-5t^3+4t)$$

which is easy to verify is equivariant with respect to the above involution, as all the terms in the polynomial are odd.