TITLE: How many Latin Squares are there with the following restrictions?
QUESTION [6 upvotes]: This problem comes from a study of game design, specifically the wonderful abstract strategy game Kamisado, which I highly recommend for the mathematically minded.  
The rules of the game are not so important, I am interested only in the board, which looks like this -

                           

The board is an 8x8 Latin Square. I wonder how much choice the designer had in choosing the board. The only restrictions we want are that the board should be the same for both players, and that swapping two colours doesn't matter in terms of gameplay. 
So the question is, more generally, "How many Latin squares of size $n$ exist which are isomorphic under a rotation of 180 degrees and symbol permutation?"
Specifically, for $n=8$, "How many possible Kamisado boards are there?". I know the answer must be in the literature somewhere, but am unfamiliar with the area and terminology. If there is a similar question somewhere please point me towards that instead.

REPLY [2 votes]: Your question pertains to the number of Latin squares with a particular symmetry ($180^{\circ}$ rotational symmetry -- so the dimension $n$ must be even), up to equivalence under permutation of labels (colors).  To choose a single representative of each such equivalence class, consider only normalized Latin squares, which are those with first row equal to $\{1,2,\ldots,n\}$.  You then want to fill in rows $2$ through ${n/2}$ in such a way that each row is a permutation of the values $1$ through $n$, and moreover, each entry $x_{i, j}$ is distinct from the entries above it in both its own column and its left-right mirror-image column: that is,
$$
x_{i,j}\not\in \left\{ x_{i,k}: k<j \right\} \cup \left\{ x_{n-i,k}: k<j \right\}
$$
for $2 \le i \le n/2$ and $1\le j \le n$.  The remaining rows are then forced by the rotational symmetry. For instance, for $n=4$ you need to fill in the second row below with $1,2,3,4$ as constrained here:
$$
\left[\begin{matrix}
1 & 2 & 3 & 4\\
\{2,3\}& \{1,4\}&\{1,4\}&\{2,3\}
\end{matrix}\right].
$$
There are clearly just $4$ ways to do this.  For $n=8$, the first two rows are:
$$
\left[\begin{matrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\
\not\in\{1,8\} & \not\in\{2,7\} & \not\in\{3,6\} & \not\in\{4,5\} 
& \not\in\{4,5\} & \not\in\{3,6\} & \not\in\{2,7\} & \not\in\{1,8\}
\end{matrix}\right].
$$
By brute force, there are $4752$ ways to fill in the first two rows.  For each of these, there are at most $4^8 2^8 = 16777216$ ways to fill in rows $3$ and $4$; so a (generous) upper bound on the number you're looking for is $7.9725330432 \times 10^{10}$.  This is quite accessible for a direct calculation.