Find all $a\in\mathbb R$ for which there exists a function $f\colon\mathbb R\rightarrow\mathbb R$, such that
(i) $f(f(x))=f(x)+x$, for all $x\in\mathbb R$,
(ii) $f(f(x)-x)=f(x)+ax$, for all $x\in\mathbb R$.
1. **Substitute \( x := f(x) \) into the second equation:**
   \[
   f(f(f(x)) - f(x)) = f(f(x)) + af(x)
   \]
   Using the first equation \( f(f(x)) = f(x) + x \), we get:
   \[
   f(f(x)) + af(x) = f(f(f(x)) - f(x))
   \]
   Since \( f(f(x)) = f(x) + x \), we can rewrite the left-hand side as:
   \[
   f(x) + x + af(x)
   \]

2. **Simplify the equation:**
   \[
   f(f(f(x)) - f(x)) = f(x) + x + af(x)
   \]
   Using the first equation again, we know:
   \[
   f(f(f(x)) - f(x)) = f(f(x) + x - f(x)) = f(x)
   \]
   Therefore, we have:
   \[
   f(x) = f(x) + x + af(x)
   \]

3. **Isolate \( f(x) \):**
   \[
   f(x) = f(x) + x + af(x)
   \]
   Subtract \( f(x) \) from both sides:
   \[
   0 = x + af(x)
   \]
   Solving for \( f(x) \):
   \[
   f(x) = -\frac{x}{a}
   \]

4. **Substitute \( f(x) = -\frac{x}{a} \) into the first equation:**
   \[
   f(f(x)) = f\left(-\frac{x}{a}\right) = -\frac{-\frac{x}{a}}{a} = \frac{x}{a^2}
   \]
   According to the first equation:
   \[
   f(f(x)) = f(x) + x
   \]
   Substitute \( f(x) = -\frac{x}{a} \):
   \[
   \frac{x}{a^2} = -\frac{x}{a} + x
   \]

5. **Solve for \( a \):**
   \[
   \frac{x}{a^2} = -\frac{x}{a} + x
   \]
   Multiply through by \( a^2 \):
   \[
   x = -ax + a^2 x
   \]
   Combine like terms:
   \[
   x = x(a^2 - a - 1)
   \]
   For non-zero \( x \):
   \[
   a^2 - a - 1 = 0
   \]
   Solve the quadratic equation:
   \[
   a = \frac{1 \pm \sqrt{5}}{2}
   \]

6. **Verification:**
   Substitute \( a = \frac{1 \pm \sqrt{5}}{2} \) back into the second equation to verify:
   \[
   f(f(x) - x) = f(x) + ax
   \]
   Using \( f(x) = -\frac{x}{a} \):
   \[
   f\left(-\frac{x}{a} - x\right) = -\frac{x}{a} + ax
   \]
   Simplify:
   \[
   f\left(-\frac{x + ax}{a}\right) = -\frac{x}{a} + ax
   \]
   \[
   f\left(-\frac{x(1 + a)}{a}\right) = -\frac{x}{a} + ax
   \]
   \[
   -\frac{-\frac{x(1 + a)}{a}}{a} = -\frac{x}{a} + ax
   \]
   \[
   \frac{x(1 + a)}{a^2} = -\frac{x}{a} + ax
   \]
   \[
   \frac{x(1 + a)}{a^2} = -\frac{x}{a} + ax
   \]
   This holds true for \( a = \frac{1 \pm \sqrt{5}}{2} \).

The final answer is \( \boxed{ a = \frac{1 \pm \sqrt{5}}{2} } \)