combinatorics - How the solve the parameter of the conjugate permutations

As we know the definition of conjugate permutations is: $$\exists p \quad p^{-1} \alpha p=\beta$$ When I have an alpha=Cycles[{{1,4},{2,5,6,3}}] and a beta=Cycles[{{1,2,5,3},{4,6}}]. So how to use Mathematica to solve the $p$?

Answer

The theoretical work from This post.

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Happy to show my own finP.And I'm glad to seen another better solution can do this all the same. :)

findP[Cycles[c1_], Cycles[c2_]] := 
 Module[{l}, l = Sort /@ {c1, c2}; 

  PermutationCycles /@ 
   Map[Last, 
    Union[Transpose[Catenate /@ l], #] & /@ 
     Function[list, 
       Transpose[{First[list], #}] & /@ Permutations[Last[list]]][
      Complement[Range[Max[l]], Flatten[#]] & /@ l], {2}]]

$\color{blue}{\text{First example}}$

findP[Cycles[{{1, 4}, {2, 5, 6, 3}}], Cycles[{{1, 2, 5, 3}, {4, 6}}]]

{Cycles[{{1,4,6,5,2}}]}

verification

PermutationProduct[InversePermutation[Cycles[{{1, 4, 6, 5, 2}}]], 
 Cycles[{{1, 4}, {2, 5, 6, 3}}], Cycles[{{1, 4, 6, 5, 2}}]]

Cycles[{{1, 2, 5, 3}, {4, 6}}]

$\color{blue}{\text{Second example}}$

twoP=findP[Cycles[{{1,3},{4,7,6}}],Cycles[{{1,5},{2,6,4}}]]

We get two $p$

{Cycles[{{2,3,5,7,6,4}}],Cycles[{{2,7,6,4},{3,5}}]}

verification

PermutationProduct[InversePermutation[#],Cycles[{{1,3},{4,7,6}}],#]&/@twoP

{Cycles[{{1,5},{2,6,4}}],Cycles[{{1,5},{2,6,4}}]}