Does every $9 \times 9$ Latin square contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$? 
Q: Does every $9 \times 9$ Latin square on the symbol set $\{1,2,\ldots,9\}$ contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?

This one has $1728$ such submatrices, which is as low as I've gotten:
$$
\begin{bmatrix}
6 & 7 & 8 & 9 & 1 & 4 & 2 & 3 & 5 \\
5 & 6 & 1 & 7 & 2 & 8 & 3 & 4 & 9 \\
9 & 1 & 6 & 2 & 4 & 3 & 7 & 5 & 8 \\
4 & 5 & 3 & 6 & 8 & 7 & 1 & 9 & 2 \\
1 & 2 & 4 & 8 & 3 & 5 & 9 & 6 & 7 \\
2 & 3 & 7 & 4 & 9 & 6 & 5 & 8 & 1 \\
8 & 9 & 2 & 3 & 5 & 1 & 6 & 7 & 4 \\
7 & 8 & 5 & 1 & 6 & 9 & 4 & 2 & 3 \\
3 & 4 & 9 & 5 & 7 & 2 & 8 & 1 & 6 \\
\end{bmatrix}$$
It doesn't seem likely that random Latin squares will help much; they average in the thousands of such submatrices.  The one above is the best random Latin square I've found so far (although, I haven't busted a gut doing this; it seems like it won't work anyway).
The groups of order $9$ have lots ($C_9$ has $5832$ and $C_3 \times C_3$ has $19440$).

This question was motivated by answering this math.SE question which asks if any $9 \times 9$ Latin square can have its rows and columns permuted to give a sudoku square.
One way to find an explicit counterexample would be to find a $9 \times 9$ Latin square with no $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$.  But this attempt didn't work since I couldn't find one.  Hence my question.
 A: How about this one?
$$
\begin{bmatrix}
  1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
  2 & 1 & 4 & 3 & 6 & 7 & 8 & 9 & 5 \\
  3 & 4 & 2 & 1 & 8 & 9 & 5 & 6 & 7 \\
  4 & 3 & 1 & 2 & 7 & 8 & 9 & 5 & 6 \\
  5 & 6 & 9 & 8 & 2 & 3 & 4 & 7 & 1 \\
  6 & 7 & 5 & 9 & 1 & 2 & 3 & 4 & 8 \\
  7 & 8 & 6 & 5 & 9 & 1 & 2 & 3 & 4 \\
  8 & 9 & 7 & 6 & 4 & 5 & 1 & 2 & 3 \\
  9 & 5 & 8 & 7 & 3 & 4 & 6 & 1 & 2 \\
\end{bmatrix}$$
I discovered it by modifying a Latin Square solver to add the restriction that no 3x3 submatrix can contain 9 different values. It found this in about 1 hour.
A: We might consider the general concept of a saturated sub-matrix, or $SSM$, being an $M \times K$ sub-matrix within a Latin square of of rank $N = M \times K$, that contains all $N$ values.
We are looking for Latin squares that are completely void of $SSM$s, so they might well be called poly-unsaturated!
For example $6 = 3 \times 2$.  Of the 9408 reduced forms, just 6 are poly-unsaturated. Here is one of them:
$$\begin{bmatrix}  1 & 2 & 3 & 4 & 5 & 6 \\
  2 & 3 & 4 & 1 & 6 & 5 \\
  3 & 1 & 5 & 6 & 2 & 4 \\
  4 & 5 & 6 & 2 & 1 & 3 \\
  5 & 6 & 1 & 3 & 4 & 2 \\
  6 & 4 & 2 & 5 & 3 & 1
\end{bmatrix}$$
The counter is still running for $9 = 3 \times 3$, but these are definitely scarce objects. 
I have also confirmed existence for $8 = 4 \times 2$. One is tempted to conjecture that for all $M,K \gt 1$, whenever $M \neq N$ they will be found to exist.
For $M = K$, well we now know that $N=9$ has poly-unsaturated forms, and $N=4$ does not. I don't know enough about cyclic groups to predict the result for $N=16$. 
