Consider a nonzero integer number $n$ and the function $f:\mathbb{N}\to \mathbb{N}$ by 
\[ f(x) = \begin{cases}
		\frac{x}{2} & \text{if } x \text{ is even} \\
		\frac{x-1}{2} + 2^{n-1} & \text{if } x \text{ is odd}
\end{cases}. 
\] Determine the set: \[ 
	A = \{ x\in \mathbb{N} \mid \underbrace{\left( f\circ f\circ ....\circ f \right)}_{n\ f\text{'s}}\left( x \right)=x \}. 
	\]
1. **Binary Representation and Function Behavior**:
   - Write \( x \) in its binary representation.
   - The function \( f(x) \) operates as follows:
     \[
     f(x) = \begin{cases}
     \frac{x}{2} & \text{if } x \text{ is even} \\
     \frac{x-1}{2} + 2^{n-1} & \text{if } x \text{ is odd}
     \end{cases}
     \]
   - For an even \( x \), \( f(x) \) simply shifts the binary digits of \( x \) to the right by one place.
   - For an odd \( x \), \( f(x) \) shifts the binary digits of \( x \) to the right by one place and adds \( 2^{n-1} \) to the result, effectively moving the last digit to the first place.

2. **Effect of \( f \) on Binary Digits**:
   - Applying \( f \) repeatedly \( n \) times will cycle the binary digits of \( x \) such that the last digit of \( x \) is moved to the first place, and the remaining digits are shifted right.
   - After \( n \) applications of \( f \), the binary representation of \( x \) will return to its original form if and only if \( x \) has \( \leq n \) binary digits.

3. **Set \( A \) Determination**:
   - The set \( A \) consists of all \( x \in \mathbb{N} \) such that applying \( f \) \( n \) times returns \( x \) to its original value.
   - This implies that \( x \) must have \( \leq n \) binary digits.
   - The maximum value of \( x \) with \( n \) binary digits is \( 2^n - 1 \).

4. **Conclusion**:
   - Therefore, the set \( A \) is:
     \[
     A = \{ x \in \mathbb{N} \mid 1 \leq x \leq 2^n \}
     \]

The final answer is \(\boxed{A = \{1, 2, \dots, 2^n\}}\).