First note that $n \equiv 1 \pmod{19}$ implies $n^5 + n^4 + n^3 + n^2 + n + 1 \equiv 6 \pmod{19}$. Assume now that $n \not\equiv 1 \pmod{19}$, i.e. $\gcd(19, n-1) = 1$. Because $n^6 - 1 = (n-1)(n^5 + n^4 + n^3 + n^2 + n + 1)$ we then have
$$19 \mid n^5 + n^4 + n^3 + n^2 + n + 1 \iff n^6 \equiv 1 \pmod{19}.$$
Because 19 is a prime number, the congruence $a^2 \equiv 1 \pmod{19}$ has exactly two solutions, namely $a \equiv \pm 1 \pmod{19}$. This is so because $(a-1)(a+1)$ can only be divisible by the prime number 19 if one of the two factors is so. This shows that $n^6 \equiv 1 \pmod{19}$ iff $n^3 \equiv \pm 1 \pmod{19}$.
To find all such $n$ we create the following table, in which we first calculated $n^2 \pmod{19}$ to keep the numbers small.