The two sequences $(1, 2, 3, \ldots)$ and $(3, 2, 1, 4, 5, 6, \ldots)$.

For a positive integer $n$, let $S_n := \{a_1, \dots, a_n\}$.

Now fix $n$ and let $M := \max S_n$ and $m := \min S_n$ and let $k := M - m$. Since all elements of $S_n$ are different by (ii), we have $k \ge n-1$. If $k \ge n$ then $M, m$ are elements of $S_k$ and $M \equiv m \pmod k$, which contradicts (ii). Hence $k = n-1$ and so the $n$ numbers in $S_n$ are $n$ consecutive positive integers. It follows that for any $n \ge m \ge 1$, we have $a_n - a_m \le n-1$. Moreover, it is easy to see that if $S_N = \{1, 2, \dots, N\}$ for some $N$, then $a_n = n$ for all $n \ge N+1$.

(i) If $a_2 = 1$ then $a_1 = 2$, since elements of $S_2$ are consecutive. But $a_4 = 2a_2 = 2$, a contradiction.

(ii) If $a_2 = 2$ and $a_1 = 1$, then $a_n = n$ for each $n \ge 3$. This gives the first answer.

(iii) If $a_2 = 2$ and $a_1 = 3$, then $a_4 = 4$ and $a_3 = 1$. Then $a_n = n$ for each $n \ge 5$ and this gives the second answer.