A nice explanation of the algorithm; thanks. However, this part isn't quite right.

>  # @p is the position of each element in @o,
>  # that is, @o = (0..$#e)[@p]
>  my @p= (0..$#e);
>  # Note that it is also true that @p = (0..$#e)[@o].
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In the language of group theory, (0..$#e) is the identity, while @o, @p are inverses of each other. Thus @p[@o]=@o[@p]=(0..$#e). For example,

  # an arbitrary permutation
  @o = @permutation = (1,2,0);
  # position of 0 is in $o[2], so $p[0]=2.
  # This leads to
  @p = @positions   = (2,0,1); 
  # And then 
  print " o[p] = (@o[@p]) = p[o] = (@p[@o]) \n";
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which outputs

 o[p] = (0 1 2) = p[o] = (0 1 2)
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Regards,

 barrachois