Let $P$ be a polyhedron which satisfies the following three conditions:


1. $P$ is built out of regular hexagons and regular pentagons.
2. Three faces meet at each vertex.
3. $P$ 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?



The Symmetry of a Soccer Ball
Only soccer ball or dodecahedron.




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Clearly 3 hexagons can not meet at one vertex.
Thus we have only 3 choices for one vertex:


* 3 pentagons
* 2 pentagons + 1 hexagon
* 1 pentagons + 2 hexagon


Note that if $[pq]$ is an edge then $p$ has the same type as $q$ (the type is determined by angle at $[pq]$). 
Thus the polyhedron is completely determined by one vertex. 
Further:


* Once you have a vertex of the first type you have a regular dodecahedron.
* If you have a vertex of the second type then you will get one hexagon surrounded by pentagons. Then it is easy to see that you can not continue.
* For the third type you will get a soccer ball or "truncated icosahedron" as some people call it :)


