9. Let us number the vertices, starting from $S$ and moving clockwise. In that case $S=1$ and $F=5$. After an odd number of moves to a neighboring point we can be only on an even point, and hence it follows that $a_{2 n-1}=0$ for all $n \in \mathbb{N}$. Let us define respectively $z_{n}$ and $w_{n}$ as the number of paths from $S$ to $S$ in $2 n$ moves and the number of paths from $S$ to points 3 and 7 in $2 n$ moves. We easily derive the following recurrence relations: $$a_{2 n+2}=w_{n}, \quad w_{n+1}=2 w_{n}+2 z_{n}, \quad z_{n+1}=2 z_{n}+w_{n}, \quad n=0,1,2, \ldots$$ By subtracting the second equation from the third we get $z_{n+1}=w_{n+1}-$ $w_{n}$. By plugging this equation into the formula for $w_{n+2}$ we get $w_{n+2}-$ $4 w_{n+1}+2 w_{n}=0$. The roots of the characteristic equation $r^{2}-4 r+2=0$ are $x=2+\sqrt{2}$ and $y=2-\sqrt{2}$. From the conditions $w_{0}=0$ and $w_{1}=2$ we easily obtain $a_{2 n}=w_{n-1}=\left(x^{n-1}-y^{n-1}\right) / \sqrt{2}$.