Find all functions $ f: \mathbb{Q}^{\plus{}} \rightarrow \mathbb{Q}^{\plus{}}$ which satisfy the conditions:

$ (i)$ $ f(x\plus{}1)\equal{}f(x)\plus{}1$ for all $ x \in \mathbb{Q}^{\plus{}}$

$ (ii)$ $ f(x^2)\equal{}f(x)^2$ for all $ x \in \mathbb{Q}^{\plus{}}$.
1. **Given Conditions:**
   - \( f(x + 1) = f(x) + 1 \) for all \( x \in \mathbb{Q}^+ \)
   - \( f(x^2) = f(x)^2 \) for all \( x \in \mathbb{Q}^+ \)

2. **Induction on \( f(x + n) \):**
   - Base case: For \( n = 1 \), \( f(x + 1) = f(x) + 1 \).
   - Inductive step: Assume \( f(x + k) = f(x) + k \) for some \( k \in \mathbb{N} \). Then,
     \[
     f(x + (k + 1)) = f((x + k) + 1) = f(x + k) + 1 = (f(x) + k) + 1 = f(x) + (k + 1).
     \]
   - By induction, \( f(x + n) = f(x) + n \) for all \( n \in \mathbb{N} \).

3. **Analyzing \( f \left( \frac{a}{b} + b \right)^2 \):**
   - For positive integers \( a \) and \( b \), consider \( f \left( \left( \frac{a}{b} + b \right)^2 \right) \):
     \[
     f \left( \left( \frac{a}{b} + b \right)^2 \right) = f \left( \frac{a^2}{b^2} + 2a + b^2 \right).
     \]
   - Using the induction result, we have:
     \[
     f \left( \frac{a^2}{b^2} + 2a + b^2 \right) = f \left( \frac{a^2}{b^2} \right) + 2a + b^2.
     \]
   - Also, from the second condition:
     \[
     f \left( \left( \frac{a}{b} + b \right)^2 \right) = f \left( \frac{a}{b} + b \right)^2.
     \]
   - Therefore:
     \[
     f \left( \frac{a}{b} + b \right)^2 = f \left( \frac{a^2}{b^2} \right) + 2a + b^2.
     \]

4. **Simplifying the Equation:**
   - Let \( y = \frac{a}{b} \). Then:
     \[
     f(y + b)^2 = f(y^2) + 2a + b^2.
     \]
   - Using the first condition:
     \[
     f(y + b) = f(y) + b.
     \]
   - Substituting this into the equation:
     \[
     (f(y) + b)^2 = f(y^2) + 2a + b^2.
     \]
   - Expanding and simplifying:
     \[
     f(y)^2 + 2bf(y) + b^2 = f(y^2) + 2a + b^2.
     \]
   - Canceling \( b^2 \) from both sides:
     \[
     f(y)^2 + 2bf(y) = f(y^2) + 2a.
     \]

5. **Solving for \( f(y) \):**
   - Since \( y = \frac{a}{b} \), we can rewrite \( 2a \) as \( 2by \):
     \[
     f(y)^2 + 2bf(y) = f(y^2) + 2by.
     \]
   - For \( y = \frac{a}{b} \), we have:
     \[
     f \left( \frac{a}{b} \right)^2 + 2b f \left( \frac{a}{b} \right) = f \left( \left( \frac{a}{b} \right)^2 \right) + 2a.
     \]
   - This implies:
     \[
     f \left( \frac{a}{b} \right) = \frac{a}{b}.
     \]

6. **Conclusion:**
   - Since \( f \left( \frac{a}{b} \right) = \frac{a}{b} \) for all positive rational numbers \( \frac{a}{b} \), we conclude that:
     \[
     f(x) = x \quad \forall x \in \mathbb{Q}^+.
     \]

The final answer is \( \boxed{ f(x) = x } \) for all \( x \in \mathbb{Q}^+ \).