Determine the real number $a$ having the property that $f(a)=a$ is a relative minimum of $f(x)=x^4-x^3-x^2+ax+1$.
To determine the real number \( a \) such that \( f(a) = a \) is a relative minimum of \( f(x) = x^4 - x^3 - x^2 + ax + 1 \), we need to follow these steps:

1. **Find the first derivative of \( f(x) \):**
   \[
   f'(x) = \frac{d}{dx}(x^4 - x^3 - x^2 + ax + 1) = 4x^3 - 3x^2 - 2x + a
   \]

2. **Set the first derivative equal to zero to find critical points:**
   \[
   f'(x) = 0 \implies 4x^3 - 3x^2 - 2x + a = 0
   \]

3. **Since \( f(a) = a \) is a relative minimum, \( a \) must be a root of \( f'(x) = 0 \):**
   \[
   4a^3 - 3a^2 - 2a + a = 0 \implies 4a^3 - 3a^2 - a = 0 \implies a(4a^2 - 3a - 1) = 0
   \]
   This gives us the potential solutions:
   \[
   a = 0, \quad a = 1, \quad a = -\frac{1}{4}
   \]

4. **Check which of these values satisfy \( f(a) = a \):**
   \[
   f(0) = 0^4 - 0^3 - 0^2 + a \cdot 0 + 1 = 1 \quad (\text{not } 0)
   \]
   \[
   f(1) = 1^4 - 1^3 - 1^2 + a \cdot 1 + 1 = 1 - 1 - 1 + a + 1 = a \quad (\text{true if } a = 1)
   \]
   \[
   f\left(-\frac{1}{4}\right) = \left(-\frac{1}{4}\right)^4 - \left(-\frac{1}{4}\right)^3 - \left(-\frac{1}{4}\right)^2 + a \left(-\frac{1}{4}\right) + 1
   \]
   \[
   = \frac{1}{256} + \frac{1}{64} - \frac{1}{16} - \frac{a}{4} + 1 \quad (\text{not } -\frac{1}{4})
   \]

   Only \( a = 1 \) satisfies \( f(a) = a \).

5. **Verify that \( a = 1 \) is a relative minimum:**
   - Compute the second derivative:
     \[
     f''(x) = \frac{d}{dx}(4x^3 - 3x^2 - 2x + a) = 12x^2 - 6x - 2
     \]
   - Evaluate the second derivative at \( x = 1 \):
     \[
     f''(1) = 12(1)^2 - 6(1) - 2 = 12 - 6 - 2 = 4
     \]
     Since \( f''(1) > 0 \), \( x = 1 \) is a relative minimum.

The final answer is \( \boxed{ a = 1 } \)