A continuous function $ f(x)$ satisfies that $ \frac {d}{dx}\left(\int_0^x f(x \plus{} t)\ dt \right)\equal{} 0$. Find the necessary and sufficient condition such that $ \sum_{n \equal{} 1}^{\infty} f(2^n) \equal{} 1$.
1. Given the problem, we start with the expression:
   \[
   \frac{d}{dx}\left(\int_0^x f(x + t)\, dt \right) = 0
   \]
   Since \( f \) is continuous, it has an antiderivative \( F \) such that \( F'(x) = f(x) \).

2. Using the Fundamental Theorem of Calculus, we can rewrite the integral:
   \[
   \int_0^x f(x + t)\, dt = F(x + t) \Big|_{0}^{x} = F(2x) - F(x)
   \]

3. Taking the derivative with respect to \( x \):
   \[
   \frac{d}{dx} (F(2x) - F(x)) = 0
   \]

4. Applying the chain rule:
   \[
   \frac{d}{dx} F(2x) - \frac{d}{dx} F(x) = 0 \implies 2f(2x) - f(x) = 0
   \]

5. Solving for \( f(2x) \):
   \[
   2f(2x) = f(x) \implies f(2x) = \frac{1}{2} f(x)
   \]

6. Letting \( x = 2^n \), we get:
   \[
   f(2^{n+1}) = \frac{1}{2} f(2^n)
   \]

7. Considering the series \( \sum_{n=1}^{\infty} f(2^n) = 1 \), we start with:
   \[
   f(2) + f(2^2) + f(2^3) + \ldots = 1
   \]

8. Using the relationship \( f(2^{n+1}) = \frac{1}{2} f(2^n) \), we can express the series as:
   \[
   f(2) + \frac{1}{2} f(2) + \frac{1}{2^2} f(2) + \ldots = 1
   \]

9. Recognizing this as a geometric series with the first term \( a = f(2) \) and common ratio \( r = \frac{1}{2} \):
   \[
   f(2) \left(1 + \frac{1}{2} + \frac{1}{2^2} + \ldots \right) = 1
   \]

10. The sum of the infinite geometric series is:
    \[
    \sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k = \frac{1}{1 - \frac{1}{2}} = 2
    \]

11. Therefore:
    \[
    f(2) \cdot 2 = 1 \implies f(2) = \frac{1}{2}
    \]

Thus, the necessary and sufficient condition for \( \sum_{n=1}^{\infty} f(2^n) = 1 \) is \( f(2) = \frac{1}{2} \).

The final answer is \( \boxed{ f(2) = \frac{1}{2} } \)