Let $P_1$, $P_2$, $P_3$ be $3$ distinct odd primes, and $n>1$, such that:
$$nP_1 \equiv P3 \pmod{P_2},\ nP_2 \equiv P_1 \pmod{P_3},\ nP_3 \equiv P_2 \pmod {P_1},$$ Take $n = 2$.
Question: Is there a triple $(P_1,P_2,P_3)$ satisfying the above $3$ equations?
Comment : I noticed that there are solutions $(P_1,P_2,P_3)$ if $n>2$ in general. But I could not find one triple with $n = 2$. Any comment will be helpful. Thank you.
There are no triples with $n = 2$. First, consider if $P_2$ is the largest prime. Your first congruence relation gives that, for some $k \in \mathbb{Z}$,
$$2P_1 \equiv P_3 \pmod{P_2} \implies 2P_1 - P_3 = kP_2 \tag{1}\label{eq1A}$$
Since $P_1 \lt P_2 \implies 2P_1 \lt 2P_2$, then $k \lt 2$. Also, $P_3 \lt P_2 \implies -P_3 \gt -P_2$, so $k \gt -1$. Thus, $k \in \{0, 1\}$. With $k = 0$, this gives $2P_1 = P_3$, which is not allowed. Thus, we must have $k = 1$ which gives
$$2P_1 - P_3 = P_2 \implies 4P_1 - 2P_2 = 2P_3 \implies -2P_2 \equiv 2P_3 \pmod{P_1} \tag{2}\label{eq2A}$$
Your third congruence equation then gives
$$2P_3 \equiv P_2 \pmod{P_1} \implies -2P_2 \equiv P_2 \pmod{P_1} \implies 3P_2 \equiv 0 \pmod{P_1} \tag{3}\label{eq3A}$$
Since the primes are distinct, this means $P_1 = 3$. The left side of \eqref{eq2A} gives
$$2P_1 = 6 = P_3 + P_2 \tag{4}\label{eq4A}$$
However, since $3$ is the smallest odd prime and the primes are distinct, then $P_2 \gt 3$ and $P_3 \gt 3$, so \eqref{eq4A} is not possible.
With $P_3$ being the largest prime instead, using similar arguments as before, your second congruence equation gives
$$2P_2 - P_1 = P_3 \implies 4P_2 - 2P_1 = 2P_3 \implies -2P_1 \equiv 2P_3 \pmod{P_2} \tag{5}\label{eq5A}$$
Thus, your first congruence equation gives
$$2P_1 \equiv P_3 \pmod{P_2} \implies -2P_3 \equiv P_3 \pmod{P_2} \implies 3P_3 \equiv 0 \pmod{P_2} \tag{6}\label{eq6A}$$
Thus, as done before, this means $P_2 = 3$, with the first part of \eqref{eq5A} giving a contradiction.
Finally, consider the case where $P_1$ is your largest prime. As done before, your third congruence equation results in
$$2P_3 - P_2 = P_1 \implies 4P_3 - 2P_2 = 2P_1 \implies -2P_2 \equiv 2P_1 \pmod{P_3} \tag{7}\label{eq7A}$$
Your second congruence equation now becomes
$$2P_2 \equiv P_1 \pmod{P_3} \implies -2P_1 \equiv P_1 \pmod{P_3} \implies 3P_1 \equiv 0 \pmod{P_3} \tag{8}\label{eq8A}$$
As was done earlier, this results in $P_3 = 3$, so the first part of \eqref{eq7A} gives a contradiction.
This covers all possibilities, showing there are no solutions for $n = 2$.