It's not an original answer, it's just a supplement to the answer of `thorimur`.

    g1 = Graph[(Sort /@ 
          Flatten[Map[
            Thread[#[[1]] \[UndirectedEdge] #[[2]]] &, {{1, {2, 3, 4, 
               5}}, {2, {1, 3, 5, 6}}, {3, {1, 4, 2, 6}}, {4, {1, 3, 5, 
               6}}, {5, {1, 2, 4, 6}}, {6, {2, 3, 4, 5}}}]]) // 
        DeleteDuplicates, 
       VertexLabels -> "Name"];(*Adjacency between faces*)
    
    UnrestrictedColoringQ[g_, coloring_Association] := 
      ContainsExactly[VertexList[g], Keys[coloring]];
    
    ColoringQ[g_, c_Association] := 
      FreeQ[Map[c, EdgeList[g], {2}], v_ \[UndirectedEdge] v_, 1] /; 
       UnrestrictedColoringQ[g, c];
    
    AllUnrestrictedColorings[g_] := 
      With[{vs = VertexList[g]}, 
       AssociationThread[vs, #] & /@ 
        Tuples[Table[i, {i, Length[vs]}], Length[vs]]];
    
    AllLabeledColorings[g_] := 
      Select[AllUnrestrictedColorings[g], ColoringQ[g, #] &];
    
    AutMod[g_, clist : {___Association}, autg_List : Null] := 
     With[{AutG = 
        Replace[autg, Null :> GroupElements[GraphAutomorphismGroup[g]]]}, 
      DeleteDuplicates[
       Function[c, 
         Function[h, 
           KeySort@KeyMap[Function[v, PermutationReplace[v, h]], c]] /@ 
          AutG] /@ clist, ContainsExactly]]
    
    
    (*With g1 as above:*)
    H = GroupElements@
       PermutationGroup[{Cycles[{{2, 3, 4, 5}}], Cycles[{{1, 2, 6, 4}}]}];
    G1 = GroupElements@
       FiniteGroupData["Octahedral", "PermutationGroupRepresentation"];
    G2 = GroupElements@
       PermutationGroup[{Cycles[{{2, 3, 4, 5}}], Cycles[{{1, 2, 6, 4}}], 
         Cycles[{{1, 6}}]}];
    num1 = AutMod[g1, AllLabeledColorings[g1]] // Length
    num2 = AutMod[g1, AllLabeledColorings[g1], G1] // Length
    num3 = AutMod[g1, AllLabeledColorings[g1], G2] // Length
    num4 = AutMod[g1, AllLabeledColorings[g1], H] // Length

The above code takes about 800 seconds to calculate `num2`. And the results of the above codes are `215`, `1860`, `215`, `230`.

Where `num1` = `num3`, this conclusion is very useful. But one thing I'm confused about is that groups `G1` and `G2` are both groups of order `48`, representing regular hexahedral groups.