Not too sure I fully understand.
I think the idea is filling triangular faces with smaller triangles, 6 along each edge (example used in the wiki)
So creating a 2D approximation, flattening out the icosahedron we get a net of 20 triangular faces, 10 squares in a 5x5 box

we can get the pattern in 2D

Then wrap it around a sphere (using the code from this post)

n = 5; Table[ Table[{{g, 0 + p}, {g, 1/6 + p}, {g + 1/6, 0 + p}}, {g, 0, (1 - 1/6)*n, 1/6}], {p, 0, 1*n, 1/6} ]; i0 = Show[ Graphics[{EdgeForm[Black], {Hue[RandomReal[]], Triangle[#]}}] & /@ %[[#]] & /@ Range[Length[%]]] {width, height} = ImageDimensions[i0]; w = 40; h = 45; pic = ImageTake[i0, {h, height - h}, {w, width - w}]; ParametricPlot3D[{Cos[u] Sin[v], Sin[u] Sin[v], Cos[v]}, {u, 0, 2 Pi}, {v, 0, Pi}, Mesh -> None, PlotPoints -> 100, TextureCoordinateFunction -> ({#4, 1 - #5} &), Boxed -> False, PlotStyle -> Texture[Show[i0, ImageSize -> 1000]], Lighting -> "Neutral", Axes -> False, RotationAction -> "Clip", ViewPoint -> {-2.026774, 2.07922, 1.73753418}, ImageSize -> 600]
You can adjust the shape for truncated icosahedron. Or are you are looking for a manipulation in 3D?
For some reason SpherePoints[] is removed from my mathematica, however
Graphics3D[Line[SpherePoints[100]]]
if you combine this and some form of ShortestPath
SpherePoints[100]; Graphics3D[Line[%[[Last[FindShortestTour[%]]]]
if you set the number of points to that of the number of points on the sphere they should be evenly spaced and the same effect.
PolyhedronData["Icosahedron"]? $\endgroup$