What if a twin is in a rocket in a synchronous orbit with the equator of the Earth while the second twin is standing at the equator watching his twin in space. If we use 0.87c for ease of calculation then the orbit needs only be about 1/3 of a light year out. All frames of reference being equal, the twins would have to think they are stationary to each other though one is traveling at 0.87c. How would this theory play out then? Even their light clocks would appear the same. Now place another non-rotating observer at the North Pole observing the twins and a real paradox develops. Since the twins experience the same referential frame they would see each other age the same. The North Pole observer should witness a 2:1 ratio between them. Run it for ten years and then have all three stand together and argue over which of the twins is older. Solutions?
All frames of reference being equal,
All frames of reference are not equal and, in this case, they're especially unequal. If my calculation is correct, for the rocket to travel around Earth at $0.87c$ at a distance of 1/3 light-year requires a proper inward acceleration of about $2,669\,\mathrm g$. The twin on the Earth has a proper acceleration of about $1\mathrm g$.
A result from special relativity is that accelerated clocks run slower than unaccelerated clocks.
From the Wikipedia article "Time Dilation":
From the local frame of reference (the blue clock), a relatively accelerated (e.g. red) clock moves more slowly