Let $f = \sum_{n=0}^{\infty} a_n e_n $ where $e_n$ are an ONB of $L^2[0,1].$
Now assume we have that $$\frac{d}{dx}e_n = \lambda_n e_n.$$
Assume $f \in H^1[0,1],$ so i.e. $||f'||_{L^2} < \infty$
I want to say now that $$\frac{d}{dx} f = \sum_{n} a_n \lambda_n e_n$$ but I am not sure how to justify the interchange of summation and differentiation.
So I am trying to make rigorous here what physicists always do.
The following answers your question, but makes some additional assumptions (see below).
Since $f' \in L^2$, and since $(e_n)_n$ is an ONB of $L^2$, we know
$$ f' = \sum_n \langle f', e_n\rangle e_n. $$
So your claim is equivalent to $\langle f', e_n \rangle = \lambda_n a_n$ for all $n$. By using partial integration, we get
\begin{eqnarray*} \langle f', e_n \rangle &=& \int_0^1 f'(x) \overline{e_n (x)} \, dx \\ &=& [f \cdot \overline{e_n}] \bigg|_0^1 - \int f(x) \overline{e_n ' (x)} \, dx \\ &=& [f \cdot \overline{e_n}] \bigg|_0^1 - \overline{\lambda_n} \int f(x) \overline{e_n(x)} \,dx \\ &=& [f \cdot \overline{e_n}] \bigg|_0^1 - \overline{\lambda_n} a_n . \end{eqnarray*}
If we now make the additional assumption that $f$ and $e_n$ are periodic of period $1$ (or if $f(0) = f(1) = 0$ or $e_n (0) = e_n(1) = 0$), the boundary terms vanish.
In this case, your claim is equivalent to $\lambda_n = -\overline{\lambda_n}$, i.e. to the condition that $\lambda_n$ is purely imaginary. That this is the case can be checked (under the same assumptions as above) by choosing $f = e_n$.