The sequence is periodic iff it is so $\mod$ all prime powers $p^k$ dividing $n$ (for all $a$). If $p^2|n$ for some prime $p$, then the sequence $p,p^2,p^3,... \equiv p,0,0,... \mod p^2$ is not periodic, thus $n$ has to be squarefree. If $p \nmid a$, then $a^{p-1} \equiv 1 \mod p$ and it is periodic. If $p|a$, then it's periodic, too. Thus the answer: exactly the squarefree $n$.