 Solution

Let $n=p^k$ and take $G=C_{p^k}$. If it embedded in $S_{n-1}$, that symmetric group would contain a permutation of order $p^k$. The order of a permutation is the least common multiple of its cycle lengths. For this least common multiple to be divisible by $p^k$, one cycle length must itself be divisible by $p^k$, requiring at least $p^k=n$ points. This is impossible in $S_{n-1}$. Thus $C_n$ is the required group.