TITLE: The Symmetry of a Soccer Ball
QUESTION [14 upvotes]: Let $P$ be a polyhedron which satisfies the following three conditions:

$P$ is built out of regular hexagons and regular pentagons.
Three faces meet at each vertex.
$P$ is topologically a sphere.

An easy Euler characteristic argument tells you that $P$ has exactly twelve pentagonal faces.
An example of a polyhedron like this is a truncated icosahedron (soccer ball for those of us in the States, football for everyone else). In this case, the pentagonal faces are arranged with some nice symmetry, and the polyhedron has icosahedral symmetry.
Another (trivial) example is the regular dodecahedron, which again has icosahedral symmetry.
Here's my question: Is this symmetry forced? What, if anything, can be said in general about the symmetry of a polyhedron which satisfies the above three conditions? 
Edit: Since the discussion below points out that there are precisely two polyhedra which satisfy the above conditions, a suitable evolution of the question, which has already begun to be discussed below, is this: What symmetry groups can a polyhedron have if one or more of the above conditions are relaxed?

REPLY [3 votes]: If you are willing to relax the trivalency requirement (and not require convexity, which was not one of the stated constraints), you can make all sorts of polyhedra, using only regular pentagons and hexagons, which have all sorts of symmetry. 
For example, start with two truncated icosahedra, and remove one hexagon from each of them. Now you can glue them together along the removed hexagons (matching up pentagonal and hexagonal faces), forming a new polyhedron with 3 and 4-valent vertices and considerably less symmetry than what you started with. Continuing, you could make long rod-shaped polyhedra, or you could make some spiky polyhedron by replacing multiple hexagons with other truncated icosahedra.