In my opinion, the question is completely arbitrary. There is no reason to expect relationships between the dynamics of $f$ and that of its derivative. Their relationship will even change under affine conjugacy.

This being said, the answer to the question is positive in that such a function $f$ does indeed exist; i.e. the conjecture discussed in the comments below the questionis false.

Indeed, for a counterexample consider $\newcommand{\eps}{\varepsilon}f(z)=\eps e^{-z}$. Then $f'(z) = -\eps e^{-z}=-f(z)$. For sufficiently small $\eps$, the right half plane $H_R$ belongs to the basin of attraction of an attracting fixed point, for both maps. Hence the Julia sets are both contained in the left half plane $H_L$. However, clearly $f^{-1}(H_L)\cap(f')^{-1}(H_L)=f^{-1}(H_L)\cap f^{-1}(H_R)=\emptyset$. Hence $J(f)\cap J(f')=\emptyset$. In particular also $I(f)\cap I(f')=\emptyset$.

**EDIT:** If you are willing to let $f$ be a polynomial, then here is an even simpler example. Take $f(z)=z^3/3 + c$. Then $f'(z)=z^2$, so $J(f')$ is the unit circle. For large $c$, clearly the closed unit disc belongs to the basin of $\infty$ of $f$, and hence does not intersect $J(f)$. (A simple calculation shows that $c=4$ will suffice.)

Of course, the same argument will work for *any* family of the form $f(z)=p(z)+c$, with $p$ a fixed polynomial of degree at least $3$. (The latter assumption is there to ensure that dynamics of $f'$ is nontrivial.)