Find all real numbers $a$ such that there exists a function $f:\mathbb R\to \mathbb R$ such that the following conditions are simultaneously satisfied: (a) $f(f(x))=xf(x)-ax,\;\forall x\in\mathbb{R};$ (b) $f$ is not a constant function; (c) $f$ takes the value $a$. 
1. **Assume the existence of a function \( f: \mathbb{R} \to \mathbb{R} \) that satisfies the given conditions.**
   - Given conditions:
     - \( f(f(x)) = x f(x) - ax \) for all \( x \in \mathbb{R} \)
     - \( f \) is not a constant function
     - \( f \) takes the value \( a \)

2. **Let \( f(x_0) = a \) for some \( x_0 \in \mathbb{R} \).**
   - Substitute \( x = x_0 \) into the functional equation:
     \[
     f(f(x_0)) = x_0 f(x_0) - ax_0
     \]
   - Since \( f(x_0) = a \), we have:
     \[
     f(a) = x_0 a - ax_0 = 0
     \]
   - Therefore, \( f(a) = 0 \).

3. **Substitute \( x = a \) into the functional equation:**
   \[
   f(f(a)) = a f(a) - aa
   \]
   - Since \( f(a) = 0 \), we have:
     \[
     f(0) = -a^2
     \]

4. **Substitute \( x = 0 \) into the functional equation:**
   \[
   f(f(0)) = 0 f(0) - a \cdot 0
   \]
   - Since \( f(0) = -a^2 \), we have:
     \[
     f(-a^2) = 0
     \]

5. **Substitute \( x = -a^2 \) into the functional equation:**
   \[
   f(f(-a^2)) = -a^2 f(-a^2) - a(-a^2)
   \]
   - Since \( f(-a^2) = 0 \), we have:
     \[
     f(0) = a^3
     \]

6. **Equate the two expressions for \( f(0) \):**
   - From step 3, \( f(0) = -a^2 \)
   - From step 5, \( f(0) = a^3 \)
   - Therefore:
     \[
     -a^2 = a^3
     \]
   - This simplifies to:
     \[
     a^3 + a^2 = 0
     \]
   - Factoring out \( a^2 \):
     \[
     a^2(a + 1) = 0
     \]
   - Thus, \( a = 0 \) or \( a = -1 \).

7. **Verify the solutions \( a = 0 \) and \( a = -1 \):**
   - For \( a = 0 \):
     - The functional equation becomes \( f(f(x)) = x f(x) \).
     - A possible solution is \( f(x) = 1_{\{1\}}(x) \), where \( 1_{\{1\}}(x) \) is the indicator function that is 1 if \( x = 1 \) and 0 otherwise.
   - For \( a = -1 \):
     - The functional equation becomes \( f(f(x)) = x f(x) + x \).
     - A possible solution is \( f(x) = -1 + 1_{\{-1\}}(x) \), where \( 1_{\{-1\}}(x) \) is the indicator function that is 1 if \( x = -1 \) and 0 otherwise.

The final answer is \( \boxed{ a = 0 } \) or \( a = -1 \).