Find the probability such that when a polynomial in $\mathbb Z_{2027}[x]$ having degree at most $2026$ is chosen uniformly at random,
    $$x^{2027}-x | P^{k}(x) - x \iff 2021 | k $$
    (note that $2027$ is prime).
    
    Here $P^k(x)$ denotes $P$ composed with itself $k$ times.

[i]Proposed by Grant Yu[/i]
To solve this problem, we need to understand the structure of the polynomial \( P(x) \) in \(\mathbb{Z}_{2027}[x]\) and the conditions under which \( x^{2027} - x \) divides \( P^k(x) - x \) if and only if \( 2021 \mid k \). Here, \( P^k(x) \) denotes the composition of \( P(x) \) with itself \( k \) times.

1. **Understanding the Polynomial and the Field:**
   - Since \( 2027 \) is a prime number, \(\mathbb{Z}_{2027}\) is a finite field with \( 2027 \) elements.
   - A polynomial \( P(x) \) in \(\mathbb{Z}_{2027}[x]\) of degree at most \( 2026 \) can be seen as a function from \(\mathbb{Z}_{2027}\) to \(\mathbb{Z}_{2027}\).

2. **Cycle Structure of Polynomials:**
   - The condition \( x^{2027} - x \mid P^k(x) - x \) implies that \( P(x) \) must permute the elements of \(\mathbb{Z}_{2027}\) in such a way that the least common multiple (LCM) of the cycle lengths of \( P \) is \( 2021 \).
   - Since \( 2021 = 43 \cdot 47 \), the cycle lengths must be factors of \( 2021 \), i.e., \( 1, 43, 47, \) or \( 2021 \).

3. **Case Analysis of Cycle Lengths:**
   - Let \( ord(z) \) be the minimal \( k \) such that \( P^k(z) = z \). The condition implies that \( ord(z) \) exists and \( \text{lcm}_{z \in \mathbb{Z}_{2027}} ord(z) = 43 \cdot 47 \).
   - This means all cycles ( \( x \to P(x) \to \cdots \to P^m(x) = x \) ) have \( m \in \{1, 43, 47, 2021\} \).

4. **Counting the Number of Functions:**
   - Suppose there are \( a \) self-cycles (cycles of length 1), \( b \) cycles of length 43, \( c \) cycles of length 47, and \( d \) cycles of length 2021. Then we have the equation:
     \[
     a + 43b + 47c + 2021d = 2027
     \]
   - Additionally, either \( bc \ge 1 \) or \( d = 1 \).

5. **Case \( d = 1 \):**
   - If \( d = 1 \), then \( a + 43b + 47c = 6 \).
   - The number of ways to distribute \( a \) self-cycles is \( \binom{2027}{a} \cdot (2027 - a)! \).
   - The number of ways to distribute \( b \) cycles of length 43 is \( \frac{(43b)!}{b! (43!)^b} \cdot (42!)^b \).
   - The number of ways to distribute \( c \) cycles of length 47 is \( \frac{(47c)!}{c! (47!)^c} \cdot (46!)^c \).

6. **Combining the Cases:**
   - The total number of functions is the sum of:
     \[
     \frac{2027!}{a! b! c! 43^b 47^c}
     \]
     over all \( a + 43b + 47c = 2027 \).

The final answer is \(\boxed{\frac{2027!}{a! b! c! 43^b 47^c}}\)