If you are interested in constructing all 6-regular graphs on 20 vertices admitting a A5/icosahedral/dodecahedral vertex-transitive symmetry group, there is a very natural way to do this.

Let's take $G=A_5$ for example. Since $G$ will act faithfully and transitively on the 20 vertices, its action will be equivalent to the action of $G$ on the cosets of a subgroup of order 3.  (This is a basic fact in permutation group theory).

Up to conjugacy, there is only one class of subgroups of order 3 in $G$, hence, up to equivalence, there is only one transitive action of $G$ on 20 points. Now, to reconstruct all the 6-regular graphs, all you have to do is find a set of sub-orbits with total size 6 and which is closed under "pairing". In fact, if you want $G$ to act arc-transitively, then you need to find a self-paired sub-orbit of length 6. 

The same method can be applied to the icosahedral\dodecahedral groups. 

If you are not familiar with this method, the keyword is "coset graph".
(See  http://www.sztaki.hu/~schneider/Teaching/4P4/chapter3.html for example).

Moreover, 20 is small enough that, with a little more work, it is possible to find ALL the symmetric 6-valent graphs on 20 vertices.

In fact, all the vertex-transitive graphs up to 32 vertices of any valency are known!

(See http://symomega.wordpress.com/2012/02/27/there-are-677402-vertex-transitive-graphs-on-32-vertices/)

You could then simply work through this list, find all 6-valent vertex-transitive graphs on 20 vertices, filter out those which are not edge-transitive (or arc-transitive, you were not very clear) and compute their chromatic numbers!