The solution set is $n \equiv 0 \pmod{4}.$ This can be done by always taking $P_n$ as the orthocenter of $\triangle P_{n-3}P_{n-2}P_{n-1}$. To prove this is necessary, first see that $P_{n-3}P_{n-2}P_{n-1}$ will always be $30-30-120$ triangle or equilateral for all $n,$ and its sides will be parallel to that of $P_{n-3}P_{n-2}P_{n-1}.$ This is just induction. Let $S$ be set of points that may be represented as $$\left(\left(a + \frac{b}{2} \right) \cdot \frac{1}{3^n}, \left( \frac{b\sqrt{3}}{2} \right) \cdot \frac{1}{3^n} \right)$$where $a, b, n$ are some integers. Once we fix a point in $S,$ the choice of $n$ does not affect the parity of $a,b$ in its representation, so we color a point $\in S$ based on the (constant) parity of $a$ and $b$ over all of its possible representations (there are $4$ different colors). We can fix $\triangle P_1P_2P_3$ on the coordinate plane. Say, $P_1 = (0,0), P_2 = (1,0), P_3 = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).$ This way, $P_1,P_2, P_3,$ and the circumcenter of $\triangle P_1P_2P_3$ are all in $S$ and are colored differently. It's not hard to see that given any equilateral triangle or $30-30-120$ triangle with all vertices in $S$ and all colored differently, and with sides parallel to those of $P_1P_2P_3,$ its circumcenter and orthocenter are also in $S$ and are colored in the other color, which finishes (only $n \equiv 0 \pmod{4}$ will yield $P_n$ that are the same color as the circumcenter of $\triangle P_1P_2P_3$). $\blacksquare$