No! There exists a natural number $s$ such that $2^s \equiv 1 \pmod{n}$ (for example $s = \varphi(n)$ from Euler's theorem or because the power series of $2$ modulo $n$ is periodic). We have $2^{s-1} \equiv \frac{n+1}{2} \pmod{n}$ and so $x = \frac{n+1}{2} \in S_n$, but $2x = n+1 > n$ is not in $S_n$. Also, if $t \le \frac{n-1}{2}$ is of $S_n$, then $2t < n$ is also. Therefore, the smallest natural number $t$ such that $t \in S_n$ and $2t \notin S_n$, is $\frac{n+1}{2}$. Since this number is different for different $n$, we get what we asked for. $\square$