This is a really common effect observed in high-frequency circuits, and it's often the difference between a good one and a noisy one.
There are two relevant length scales in this problem: the first is the distance between the plane and the conductor, and the second is the wavelength of the electromagnetic waves being generated by the circuit. If you're coming at this from an electrical engineering perspective (which is where this kind of effect is most often studied) it's sometimes easy to forget that your circuits are all still just electric and magnetic fields and radiation. Whenever your wavelength is much longer than your circuit elements and wires, you can simply look for a low resistance straight-line path. Once your wavelength starts to be comparable or even shorter than your components (3 GHz corresponds to 10 cm wavelength in air) you have to start thinking in terms of EM waves and waveguides. At high frequencies, your conductors stop being equipotentials, but instead start to see charge bunching up dynamically and oscillating back and forth. The current path through the ground plane in your question changes because you cross over between these two regimes.
The clearest physical picture for the high frequency case is that power is being transferred by an EM wave. The conductors are basically sources of mobile charges which slosh around in response to the E and B fields and act to localize the wave in space. In the case of a thin conductor above a ground plane, this is what happens--the wave is localized between the two and follows along the wire. As the wave travels, it perturbs the electrons directly beneath the conductor more than anywhere else on the ground plane, which leads us to say that the return current runs directly beneath the top conductor. Your system becomes a type of what's called a "waveguide" at higher frequencies.