Solution:

Only $n \equiv 1 \pmod{210}$ work. Proof: we require $\gcd(n, 210) = 1$. Note that for all $p \leq 7$ the order of $n$ $(\bmod\ p)$ divides $p-1$, hence is relatively prime to any $p \leq 7$. So $n^{n} \equiv 1 \pmod{p} \Longleftrightarrow n \equiv 1 \pmod{p}$ for each of these $p$.