a Standard Young Tableau symmetry and the non-attacking rook problem.

Question

Any standard or semi-standard Young tableau T of shape $\lambda$ of n can be transformed into an other tableau T´ of $\lambda$ by an operation flip(T) = T´. This operation 'flip' consists of the following steps:

1. extend the rows of T with zero's to obtain a rectangular array with l($\lambda$) rows and $\lambda_1$ columns ;
2. reverse the row and column entries ;
3. replace each non-zero entry u by n+1-u resulting in a rectangular skew tableau
4. rectify (by jeu-de-taquin) this skew tableau to obtain the tableau T´ of $\lambda$.

Properties:

1. The operation 'flip' is an involution on the set of all SYT of $\lambda$. (proof ?)

2. The count of fixed points over all partitions of n equals:
   a(1..16) = 1, 2, 2, 6, 6, 20, 20, 76, 76, 312, 312, 1384, 1384, 6512, 6512, 32400
   (apparently equivalent to 'aerated' version of A000898 )

3. The set of counts of fixed points for the set of partitions of (2k+1) seems to equal that of partitions of 2k if partitions with zero fixed points are disregarded :
   n=2 : {1, 1}
   n=3 : {1, 0, 1}
   n=4 : {1, 1, 2, 1, 1}
   n=5 : {1, 0, 1, 2, 1, 0, 1}
   n=6 : {1, 1, 3, 2, 3, 0, 2, 3, 3, 1, 1}
   n=7 : {1, 0, 2, 3, 0, 1, 0, 3, 3, 1, 3, 0, 2, 0, 1}

4. up to n=16, the number of fixed points for all T of shape $\lambda$ contains only prime factors 2,3,5 and 7 ;

5. The count of 2-cycles seems to equal A000900; this sequence is similarly defined as (A000085(n)-A000898(int(n/2)))/2 , but then related not to SYT symmetries but to the n-by-n non-attacking rooks problem.

Question : this must be generally known (to those who know about such things). I'm looking for a literature reference, so that I can ammend the OEIS with more than just a conjecture.

Added 2016/12/30 15:15 UTC:
Conjecture: (tested up to n=16 for SYT and n=9 for SSYT):
for both SYT and SSYT of shape $\lambda$ : # of fixed points = 0 if the 2-core of $\lambda$ differs from resp. { } (=empty) for even n or {1} for odd n ;
for SYT of shape $\lambda$ : # of fixed points = Binomial(floor(n/2),|q1|)*h(q1)*h(q2)
for SSYT of shape $\lambda$ : # of fixed points =
for n even : c(q1,n/2)*c(q2,n/2) and for odd n : c(q1,(n-1)/2)*c(q2,(n+1)/2)
where h($\mu$) is the number of SYT of shape $\mu$ (hook length formula), q1 and q2 are the first and second 2-quotients of $\lambda$ and c($\mu$,k) is the count of SSYT of shape $\mu$ with largest part at most k (hook content formula).

Answer 1

The operation you describe is what is called (at least by me) forming the Schützenberger dual of a standard Young tableaux (and you can carry over the definition without problem to the semi-standard case). It is not defined here in the way Schützenberger originally defined it, though he might have done something close to this in a later paper (I don't think I came up with this myself). In any case I checked that the equivalence with this description is the essence of proposition$~$5.6 in this paper I wrote.

Then you are interested in fixed points of this operation, the self-dual standard Young tableaux. These are in bijection with standard domino tableaux (possibly leaving the square nearest the origin uncovered), as stated in proposition$~$2.3.3 of the same paper. More about domin tableaux can be found in this other paper of mine. This probably explains much of what you observed; I did not check all the details.