No, they aren't.
Each of these self-maps $f\in\{P,Q\}$ has a unique point with a unique preimage $\{0\}$. So any such conjugation $u$ has to fix $0$. For both $f$, we have $f^{\circ 2}(0)\neq 0$, so $f^{\circ 3}(0)$ has exactly two preimages: $f^{\circ 2}(0)$ and its negative. So $u$ has to map $-P^{\circ 2}(0)$ to $-Q^{\circ 2}(0)$. But then one sees that $-P^{\circ 2}(0)$ has two preimages by $P$ while $-Q^{\circ 2}(0)$ has no preimage by $Q$. So there exists no such conjugation.
Precisely: $-P^{\circ 2}(0)=-2$, so $P^{-1}(\{-P^{\circ 2}(0)\})$ is the set of square roots of $-3$, while $-Q^{\circ 2}(0)=-6$, so $Q^{-1}(\{-Q^{\circ 2}(0)\})$ is the set of square roots of $-8$.
Now it's a game using quadratic reciprocity:
- $p$ is 3 mod 4, so $-1$ is not a square mod $p$
- $p$ is $\pm 1$ mod 8, so $2$ is a square mod $p$.
- given this, we see that $-8$ is not a square mod $p$
- since both $3$ and $p$ are $-1$ mod 4, $3$ is a square mod $p$ iff $p$ is not a square mod $3$, i.e., if $p$ is $-1$ mod $3$, which is not the case. So $3$ is not a square mod $p$
- hence $-3$ is a square mod $p$.
Note that we reach the same conclusion for every $p$ which is $-1$$7$ mod $24$.