Find all functions $ f : Z\rightarrow Z$ for which we have $ f (0) \equal{} 1$ and $ f ( f (n)) \equal{} f ( f (n\plus{}2)\plus{}2) \equal{} n$, for every natural number $ n$.
1. **Given Conditions:**
   - \( f : \mathbb{Z} \rightarrow \mathbb{Z} \)
   - \( f(0) = 1 \)
   - \( f(f(n)) = n \)
   - \( f(f(n+2) + 2) = n \)

2. **Substitute \( n = 0 \) in the second condition:**
   \[
   f(f(0)) = 0
   \]
   Since \( f(0) = 1 \), we have:
   \[
   f(1) = 0
   \]

3. **Substitute \( n = f(n) \) in the second condition:**
   \[
   f(f(f(n))) = f(n)
   \]
   Using \( f(f(n)) = n \), we get:
   \[
   f(n) = f(n+2) + 2
   \]

4. **Solve the recurrence relation \( f(n) = f(n+2) + 2 \):**
   - Let \( f(n) = a_n \). Then, \( a_n = a_{n+2} + 2 \).
   - This is a linear recurrence relation with constant coefficients.

5. **Find the general solution of the recurrence relation:**
   - Assume \( a_n = c_1 + c_2 n \).
   - Substitute into the recurrence relation:
     \[
     c_1 + c_2 n = c_1 + c_2 (n+2) + 2
     \]
     \[
     c_1 + c_2 n = c_1 + c_2 n + 2c_2 + 2
     \]
     \[
     0 = 2c_2 + 2
     \]
     \[
     c_2 = -1
     \]
   - Therefore, \( a_n = c_1 - n \).

6. **Determine \( c_1 \) using the initial condition \( f(0) = 1 \):**
   \[
   f(0) = c_1 - 0 = 1
   \]
   \[
   c_1 = 1
   \]

7. **Thus, the function \( f(n) \) is:**
   \[
   f(n) = 1 - n
   \]

8. **Verify the solution:**
   - Check \( f(f(n)) = n \):
     \[
     f(f(n)) = f(1 - n) = 1 - (1 - n) = n
     \]
   - Check \( f(f(n+2) + 2) = n \):
     \[
     f(n+2) = 1 - (n+2) = -n - 1
     \]
     \[
     f(f(n+2) + 2) = f(-n - 1 + 2) = f(1 - n) = n
     \]

Since both conditions are satisfied, the function \( f(n) = 1 - n \) is indeed a solution.

The final answer is \( \boxed{ f(n) = 1 - n } \).