The answer: $P(x) = x, P(x) = 1, 0, -1$.

We start with the case when $\deg(P(x)) \ge 2$. Let us take $n = q$, where $q$ is prime: $P(q+d(q)) = q+d(P(q))$ yields $P(2q) = q+d(P(q))$. Therefore, $|P(2q)| \le q + |P(q)|$ and $\left|\frac{P(2q)}{P(q)}\right| \le \frac{q}{|P(q)|} + 1$ (1). Now when $q$ increases the left hand side of (1) goes to $2^{\deg(P(x))}$, but right hand side goes to 1. Contradiction.

Now let $\deg(P(x)) = 1$ and $P(x) = bx + c$. Then again for $n = q$ we get $2bq + c = q + d(bq + c)$ and $(2b - 1)q + c = d(bq + c)$. If $q$ is sufficiently large we get that $b \ge 1$ and $(2b - 1)q + c \le bq + c$ which in turn yields $b = 1$. Thus, $n + d(n) + c = n + d(n + c)$ and $d(n) + c = d(n + c)$ (2).

If $c > 0$ then for $n = 2^l - c$ the left hand side of (2) is at least 3, while the right hand side of (2) is 2. If $c < 0$ then for $n = 2^l$ the left hand side of (2) is at most 1, while the right hand side of (2) is at least 2. Thus, $c = 0$ and $P(x) = x$.

If $\deg(P(x)) = 0$ then for $c \ne 0, \pm 1$ we get $c = n + d(c)$, a contradiction.