Number of nonequivalent ways to place n non attacking rooks on a n X n board up to rotations and reflections with no rook on 2 main diagonals .
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Let a(n) be the number of nonequivalent solutions.
Let F(n) = A003471(n) be the number of solutions counting each rotation or reflection separately. 

Symmetries can be rotational (quarter turn or half turn) or a reflection on a diagonal. 
For n > 1, a horizontal or vertical line of symmetry is not possible since this would require
multiple rooks in the same row or column.

Case n odd:
  Since there are n rooks and n is odd, symmetry would require a rook to be placed in the central square
  which lies on the two main diagonals. No symmetrical solutions are therefore possible.
  Symbolically:

       a(2n+1) = (1/8) * F(2n+1).

Case n even:
  For a 2n X 2n board, let C(n), R(n) and D(n) be the number of solutions with respectively half turn
  rotational symmetry, quarter turn rotational symmetry and diagonal symmetry on a given diagonal. Then applying
  Burnside's lemma:
                
       a(2n) = (1/8) * ( F(2n) + C(n) + 2*R(n) + 2*D(n) ).

It remains then to determine D(n), R(n) and C(n).

Permutations:
  There is a simple bijection between permutations p on [n] and non attacking placements of n rooks on an n X n board.
  In particular, k is mapped to p(k) if the k'th column has a rook in row p(k).
  A board without rooks in the main diagonals corresponds to a permutation with p(k) <> k and p(k) <> n - 1 - k.
  A permutation with p(k) <> k is called a derangement or permutation without fixed points. (see A000166)

D(n): number of solutions on a 2n X 2n board invariant under a given diagonal symmetry.
  Permutations p on [2n] with diagonal symmetry have p(p(k)) = k. These are just involutions. Without p(k) = k these are perfect matchings
  in the 2n complete graph and without p(k) = 2n - 1 - k these are perfect matchings in the cocktail party graph. These
  types of permutation are counted by A053871(n).

       D(n) = A053871(n).


R(n): number of solutions on a 2n X 2n board invariant under rotation by 90 degrees.
  Permutations p on [2n] with rotational symmetry have p(p(k)) = 2n - 1 - k. 
  In general, every rotationally symmetric non attacking placement of 2n rooks cannot have a rook on the main diagonal.
  (p(k)=k implies k = p(p(k)) = 2n - 1 - k which is not possible and p(2n-1-k)=k implies p(k) = p(p(2n-1-k)) = k which is not possible.)

       R(n) = A037224(2n).

C(n): number of solutions on a 2n X 2n board invariant under rotation by 180 degrees.
  Permutations p on [2n] with diagonal symmetry have 2n - 1 - p(k) = p(2n - 1 - k).  

  Example:
        1 2 3 4 5 6
        3 5 6 1 2 4          
   Every such permutation can be decomposed uniquely into a permutation without fixed points on [n] and a binary word of length n.
   The permutation is constructed by mapping k to min(p(k), p(2n-1-k)) and the binary word is constructed
   by mapping k to 0 or 1 depending on whether p(k) or p(2n-1-k) was the least.
   
   For example, the above permutation decomposes into:
     k            1 2 3         
     permutation  3 2 1         
     bin word     0 1 1        

   Conversely, a permutation without fixed points and binary word can be used to uniquely construct a permutation p
   with the required symmetry and for every k, p(k) <> k and p(k) <> n - 1 - k.

   The number permutations without fixed points is given by A000166(n) and the number of binary words by 2^n.

         C(n) = 2^n * A000166(n).