We are tasked with finding all functions $f: \mathbb{Q} \rightarrow \mathbb{Z}$ that satisfy the functional equation:

$$f \left( \frac{f(x) + a}{b} \right) = f \left( \frac{x + a}{b} \right)$$

for all $x \in \mathbb{Q}$, $a \in \mathbb{Z}$, and $b \in \mathbb{Z}_{>0}$.

### Step 1: Consider Constant Functions

Assume that $f$ is a constant function. This means $f(x) = c$ for some fixed $c \in \mathbb{Z}$ for all $x \in \mathbb{Q}$.

Substitute $f(x) = c$ into the equation:

$$f \left( \frac{c + a}{b} \right) = f \left( \frac{x + a}{b} \right)$$

Since $f$ is constant, the left-hand side simplifies to $f \left( \frac{c + a}{b} \right) = c$. Thus, the equation holds because the right-hand side simplifies to $f \left( \frac{x + a}{b} \right) = c$ as well.

Therefore, any constant function $f(x) = c$ for $c \in \mathbb{Z}$ is a solution.

### Step 2: Consider Floor and Ceiling Functions

Next, consider the functions $f(x) = \lfloor x \rfloor$ and $f(x) = \lceil x \rceil$.

#### For $f(x) = \lfloor x \rfloor$:

Substitute into the equation:

$$\lfloor \frac{\lfloor x \rfloor + a}{b} \rfloor = \lfloor \frac{x + a}{b} \rfloor$$

The floor function generally satisfies properties that make these quantities equal because the operation of flooring "rounds down" to the nearest integer, preserving the integer status of both sides of the equation when $x$ is a rational number.

#### For $f(x) = \lceil x \rceil$:

Similarly, substitute:

$$\lceil \frac{\lceil x \rceil + a}{b} \rceil = \lceil \frac{x + a}{b} \rceil$$

The ceiling function rounds up to the nearest integer, another transformation that keeps the equality intact due to consistent rounding in both the numerator and arguments of the floor and ceiling functions for rational inputs.

### Conclusion

Thus, the solutions to the functional equation are the following functions:

$$f(x) = c \text{ for } c \in \mathbb{Z}, \quad f(x) = \lfloor x \rfloor, \quad f(x) = \lceil x \rceil.$$

The complete set of solutions can be expressed as:
$$\boxed{\{ f(x) = c \mid c \in \mathbb{Z} \} \cup \{ f(x) = \lfloor x \rfloor \} \cup \{ f(x) = \lceil x \rceil \}}$$