Since $f(f(f(s)))=s$ for all $s \in S$, each cycle in the cycle decomposition of $f$ must have length 1 or 3. Also, since $f(s) \not \equiv s \bmod 3$ for all $s \in S$, each cycle cannot contain two elements $a, b$ such that $a=b \bmod 3$. Hence each cycle has exactly three elements, one from each of residue classes mod 3. In particular, $1,4,7$ belong to distinct cycles. There are $6 \cdot 3$ ways to choose two other numbers in the cycle containing 1. Then, there are $4 \cdot 2$ ways to choose two other numbers in the cycle containing 4. Finally, there are $2 \cdot 1$ ways to choose two other numbers in the cycle containing 7. Hence the desired number of functions $f$ is $6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1=288$.