Find all prime numbers whose base $b$ representations (for some $b$) contain each of the digits $0,1,\ldots,b-1$ exactly once. (Digit $0$ may appear as the first digit.)
1. **Understanding the Problem:**
   We need to find all prime numbers whose base \( b \) representations contain each of the digits \( 0, 1, \ldots, b-1 \) exactly once. Note that digit \( 0 \) may appear as the first digit.

2. **Sum of Digits in Base \( b \):**
   The sum of the digits \( 0, 1, \ldots, b-1 \) is:
   \[
   \sum_{i=0}^{b-1} i = \frac{(b-1)b}{2}
   \]
   This sum must be congruent to the number \( p \) modulo \( b-1 \):
   \[
   p \equiv \frac{(b-1)b}{2} \pmod{b-1}
   \]

3. **Simplifying the Congruence:**
   Since \( b \equiv 1 \pmod{b-1} \), we have:
   \[
   \frac{(b-1)b}{2} \equiv \frac{(b-1) \cdot 1}{2} = \frac{b-1}{2} \pmod{b-1}
   \]
   Therefore:
   \[
   p \equiv \frac{b-1}{2} \pmod{b-1}
   \]

4. **Case Analysis Based on Parity of \( b \):**
   - **Case 1: \( b \) is even**
     \[
     b = 2k \quad \text{for some integer } k
     \]
     Then \( b-1 \) is odd, and:
     \[
     p \equiv 0 \pmod{b-1}
     \]
     For \( p \) to be a prime, \( p \) must be \( b-1 \).

   - **Case 2: \( b \) is odd**
     \[
     b = 2k+1 \quad \text{for some integer } k
     \]
     Then \( b-1 \) is even, and:
     \[
     p = k(b-1) + \frac{b-1}{2} = (2k+1) \cdot \frac{b-1}{2}
     \]
     This can only be a prime if either \( k = 0 \) or \( b = 3 \).

5. **Detailed Case Analysis:**
   - **Subcase 1: \( p = b-1 \)**
     The base \( b \) representation of \( p \) is only one digit, which does not satisfy the condition of containing all digits \( 0, 1, \ldots, b-1 \).

   - **Subcase 2: \( k = 0 \)**
     Then:
     \[
     p = \frac{b-1}{2}
     \]
     Since \( p < b \), this will also only have one digit in base \( b \), which again does not satisfy the condition.

   - **Subcase 3: \( b = 3 \)**
     Then \( p \) must be some permutation of \( 210_3 \). We can test all permutations:
     \[
     210_3 = 2 \cdot 3^2 + 1 \cdot 3 + 0 = 18 + 3 = 21
     \]
     \[
     120_3 = 1 \cdot 3^2 + 2 \cdot 3 + 0 = 9 + 6 = 15
     \]
     \[
     201_3 = 2 \cdot 3^2 + 0 \cdot 3 + 1 = 18 + 1 = 19
     \]
     \[
     102_3 = 1 \cdot 3^2 + 0 \cdot 3 + 2 = 9 + 2 = 11
     \]
     \[
     012_3 = 0 \cdot 3^2 + 1 \cdot 3 + 2 = 3 + 2 = 5
     \]
     \[
     021_3 = 0 \cdot 3^2 + 2 \cdot 3 + 1 = 6 + 1 = 7
     \]
     The prime numbers among these are \( 5, 7, 11, 19 \).

Conclusion:
\[
\boxed{5, 7, 11, 19}
\]