The Symmetry of a Soccer Ball 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?
 A: If one gives up on regularity of the hexagonal and pentagonal faces then these graphs (usually called "Fullerene graphs") don't have to have much symmetry. See e.g. this 26-vertex graph with 12 pentagonal faces, 2 hexagonal faces, and only order-four symmetry.
A: Perhaps you will also find the following interesting:
V. Braungardt und D. Kotschick: “The classification of football patterns”, 2006.
(for a popular Summary see: D. Kotschick: „The Topology and Combinatorics of Soccer Balls”, The American Scientist, Juli-August 2006)

ABSTRACT. We prove that every spherical football is a branched cover, branched only in the vertices,
  of the standard football made up of 12 pentagons and 20 hexagons. We also give examples
  showing that the corresponding result is not true for footballs of higher genera. Moreover, we
  classify the possible pairs (k; l) for which football patterns on the sphere exist satisfying a natural
  generalisation of the usual incidence relation between pentagons and hexagons to k-gons and l-gons. 

Here a football (AE: soccer ball) pattern is a map on the two-sphere satisfying
the following conditions:


*

*at least three edges meet at every vertex

*all faces are pentagons and hexagons

*the edges of each pentagon meet only edges of hexagons

*the edges of each
hexagon alternately meet edges of pentagons and of hexagons

A: From a combinatorial point of view one can define a fullerene to be a 3-valent 3-connected graph with exactly 12 faces which are 5-gons (pentagons) and h 6-gons (hexagons). By Steinitz's Theorem fullerenes which exist as graphs can be realized by convex polyhedra. Branko Grunbaum and Theodore Motzkin showed, The number of hexagons and the simplicity of geodesics on certain polyhedra, Canadian J. Math., 15 (1963) 744-751, that the admissible values of h for such graphs are all non-negative integers h except h = 1. Other proofs of this, given by construction, show other features than what Grunbaum and Motzkin did. (For references see article listed below.)
What are the symmetry groups which can arise as the automorphism groups of fullerene graphs?  There are only 28 such groups and they are listed on page 36 of the book: Geometry of Chemical Graphs: Polycylces and Two-Faced Maps, by Michel Deza, and Mathieu Dutour Sikiric, Cambridge U. Press, 2008. By a theorem of Peter Mani, these fullerene graphs can be realized by 3-dimensional polyhedra with the full automorphisms group of the graph as the group of isometries of the realizing polyhedron.
For further discussion of fullerenes and some open problems about fullerene graphs see:
Malkevitch, J., Geometrical and Combinatorial Questions about Fullerenes, in Discrete Mathematical Chemistry, (P. Hansen, P. Fowler, M. Zheng, eds.), Volume 51, DIMACS Series in Discrete Mathematics and Computer Science, AMS, Providence, 2000, pp. 261-266.
A: 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.
A: Lemma: In a polyhedron of this type with polygons $A$ and $B$ sharing an edge $e$, the two  other polygons meeting $e$ must have the same number of sides.
Proof: By local symmetry reflecting through the perpendicular bisector of $e$, the angles are equal. 
Sergei Ivanov proved the same lemma in the comments.

Since 5 is odd, all of the polygons around a pentagon must have the same number of sides, since you can't have a nonconstant alternating sequence. So, the only possibilities are that all polygons are pentagons, or that each pentagon is surrounded by hexagons, and each of these hexagons is surrounded by 3 pentagons and 3 hexagons. In the latter case, attaching pentagonal pyramids to each pentagon extends each hexagon into an equilateral triangle, producing a polyhedron whose faces are equilateral triangles with 5 meeting at a vertex, an icosahedron, so the original was a truncated icosahedron.
Note that if you have equilateral triangles and squares meeting 4 to a vertex, then there are two possibilities for 3 squares and 1 triangle at a vertex, one with cubic symmetry and one with only dihedral symmetry with a belt which is an octagonal prism. 
By contrast, if you require that there are 3 congruent triangles meeting at a vertex, but drop the regularity assumption, you get a family of disphenoids, which generically have the Klein 4-group as symmetries and no reflective symmetry. These are related to ideal hyperbolic tetrahedra.
A: Only soccer ball or dodecahedron.

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 :)

A: An interesting example involving self-intersection, but the valence is 4 is given on page 158 of Wenninger's Polyhedron Models (Model #102), this is Great Dodecahemicosahedron, with 10 hexagons and 12 pentagons. You can check it out as well on wikipedia. A similar example is the small demidecahemicosahedron. This provides a partial answer to Pisanski's follow up question.
