Solution:
First note that
$$f(x) \equiv \pm 2 x \quad \bmod 2 n+1$$
It follows that
$$f^{p}(x) \equiv \pm 2^{p} x \quad \bmod 2 n+1$$
Thus if $f^{m}(1)=1$, $2^{m} \equiv \pm 1$ and so, for any $k \in S$,
$$f^{m}(k) \equiv \pm 2^{m} k \equiv \pm k \quad \bmod 2 n+1$$
that is, $f^{m}(k) \pm k=j(2 n+1)$ for some integer $j$ and some choice of the sign. Since
$$0<1+1 \leq f^{m}(k)+k \leq n+n<2 n+1,$$
the plus sign is invalid. Thus the minus sign holds, and since
$$-(2 n+1)<1-n \leq f^{m}(k)-k \leq n-1<2 n+1,$$
we get $j=0$, i.e. $f^{m}(k)=k$.