Solution:

Apply a checkerboard coloring to the grid so that the corners and the center are black. Then by parity, the $5$ odd numbers must be in the $5$ black cells and the $4$ even numbers in the $4$ white cells.

If the center cell is $1$, there are $4$ ways to pick $2$ and $2$ ways to pick $3$, at which point the rest of the numbers are determined, for a total of $8$ possibilities. By symmetry there are $8$ possibilities for the center cell to be $9$. Likewise, if the center cell is $3$, picking $2$ and $1$ gives $8$ more, and another $8$ more for the center cell being $7$. Finally, if the center cell is $5$, the path is still determined by picking $6$ and $7$. In total, there are therefore $5 \cdot 8 = 40$ possibilities.