, 2011

Since $n$ is odd, $|S| = \phi(n) = n \prod_{p|n} (1 - 1/p)$ is even. Given an element $k$ of $S$, write
$$k^{|S|} \equiv 1 + n r_k \pmod{n^2} \quad \text{and} \quad (n-k)^{|S|} \equiv 1 + n r_{n-k} \pmod{n^2},$$
and notice that
$$(n-k)^{|S|} \equiv k^{|S|} - |S| \cdot n \cdot k^{|S|-1} \pmod{n^2},$$
to get
$$r_k - r_{n-k} \equiv |S| \cdot k^{|S|-1} \pmod{n}.$$
Hence
$$\prod_{k \in T} (r_k - r_{n-k}) \equiv |S|^{|T|} \left( \prod_{k \in T} k \right)^{|S|-1} \pmod{n}.$$
Finally, notice that the product in the right-hand member is congruent to 1 modulo $n$. To see this, let $k'$ denote the modulo $n$ multiplicative inverse of an element $k$ of $S$, and notice that
$$k' + 1 \equiv k'(k+1) \pmod{n} \quad \text{and} \quad (k-1)(k'+1) \equiv k-k' \pmod{n}.$$
The first congruence shows that if $k$ belongs to $T$, then so does $k'$, and the second shows that $k = k'$ if and only if $k = 1$. Consequently, if $k \ne 1$, the factors $k$ and $k'$ in the product can be paired off and the conclusion follows.