For the give functions in $\mathbb{N}$:
[b](a)[/b] Euler's $\phi$ function ($\phi(n)$- the number of natural numbers smaller than $n$ and coprime with $n$);
[b](b)[/b] the $\sigma$ function such that the $\sigma(n)$ is the sum of natural divisors of $n$.
solve the equation $\phi(\sigma(2^x))=2^x$.
1. **Understanding the Functions:**
   - Euler's $\phi$ function, $\phi(n)$, counts the number of integers up to $n$ that are coprime with $n$.
   - The $\sigma$ function, $\sigma(n)$, is the sum of the divisors of $n$.

2. **Given Equation:**
   We need to solve the equation $\phi(\sigma(2^x)) = 2^x$.

3. **Simplifying $\sigma(2^x)$:**
   - The divisors of $2^x$ are $1, 2, 2^2, \ldots, 2^x$.
   - Therefore, $\sigma(2^x) = 1 + 2 + 2^2 + \ldots + 2^x$.
   - This is a geometric series with the sum:
     \[
     \sigma(2^x) = \sum_{k=0}^{x} 2^k = 2^{x+1} - 1
     \]

4. **Substituting into the Equation:**
   - The equation becomes $\phi(2^{x+1} - 1) = 2^x$.

5. **Analyzing $\phi(2^{x+1} - 1)$:**
   - We need to find when $\phi(2^{x+1} - 1) = 2^x$.
   - Euler's theorem states that if $a$ and $n$ are coprime, then $a^{\phi(n)} \equiv 1 \pmod{n}$.
   - For $2^{x+1} - 1$, we need to check if $\phi(2^{x+1} - 1) = 2^x$.

6. **Divisibility Condition:**
   - From Euler's theorem, $2^{x+1} - 1 \mid 2^{\phi(2^{x+1} - 1)} - 1$.
   - This implies $x+1 \mid \phi(2^{x+1} - 1)$.
   - Given $\phi(2^{x+1} - 1) = 2^x$, we get $x+1 \mid 2^x$.

7. **Power of 2 Condition:**
   - Since $x+1 \mid 2^x$, $x+1$ must be a power of 2.
   - Let $x+1 = 2^m$, then $x = 2^m - 1$.

8. **Substituting Back:**
   - We need to check if $\phi(2^{2^m} - 1) = 2^{2^m - 1}$.
   - $2^{2^m} - 1$ is a Mersenne number, and $\phi(2^{2^m} - 1)$ can be complex to compute directly.

9. **Prime Factorization:**
   - $2^{2^m} - 1 = (2-1)(2+1)(2^2+1)\ldots(2^{2^{m-1}}+1)$.
   - Using the multiplicative property of $\phi$, we get:
     \[
     \phi(2^{2^m} - 1) = \prod \phi(2^{2^k} + 1)
     \]

10. **Checking the Product:**
    - If $\phi(n) \leq n-1$ with equality if and only if $n$ is prime, then:
      \[
      \prod \phi(2^{2^k} + 1) \leq \prod (2^{2^k} + 1 - 1) = 2^{2^m - 1}
      \]
    - This implies all $2^{2^k} + 1$ must be prime, which is not true for all $k$.

11. **Conclusion:**
    - The only solution that fits all conditions is $x = 1$.

The final answer is $\boxed{1}$