How many Latin Squares are there with the following restrictions? 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. 
 A: Some Hints:
We have to fill the first player half, then we'll have the second player's half with symetry.
Now, we have a $n$ x $ (n/2)$ table. Just randomly fill the first line, you have $n!$ alternatives, right?
Now, filling the second line. What are our restrictions? No colour can have the same column number with the previous one, of course. And the other restriction comes from the symetry. We know that if a colour is in $i^{th}$ column in $j^{th}$ line, it is also in $(n-i+1)^{th}$ column in $(n-j)^{th}$ line. Then, the other restriction is no colour can have the same column number with $(n-$previous_column_number $+1$). Then, how many alternatives do we have?
Let's start with the first one, we have n-2 alternatives. Now, assume we had put the the first colour to $i^{th}$ column in the first line and $j^{th}$ column in the second line. And now assume we had put the second colour in the first line to $j^{th}$ column. Now, for the second colour we again have n-2 alternatives. But if we had put the first colour into a column different than the one with number $j$, we would have $n-4$ alternatives for the second colour. That is the tricky part of our problem. 
Edit (further hint): What you have to find is that:
I have $n$ colours and $n$ spots. For each colour, $k$ spots are forbidden where $k$ depends on the line number, as at $i^{th}$ line, $k= n-2(i-1)$. For example, for first line there is no restriction, you can randomly locate the colours. For the second line, for each colour, 2 spots are forbidden. For third, 4 spots are forbidden etc. 
If you can find the number of permutations for that restriction, you'll have your answer. 
A: 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.
