When a clock was flown around the world,was time dilation mutual? Time dilation is always mutual - both clocks are travelling relative to each other and both must observe that the other clock is slower than themselves. Since motion is relative, and it is not possible to determine which one is noving and which one is still, so it is not possible for just one clock to be slower than the other. 
Therefore, in the famous experiment, when the atomic clock was flown around the world, was it observed and recorded that (a) the person travelling with the flying clock found that the clock on earth was slower than the flying clock, whereas (b) the person on earth found that the flying clock was slower than the clock on earth?
In particular, I am interested in the first observation, i.e. for the person travelling with the clock the earth clock was really slower than the flying clock, as that is not the result that one usually hears for this experiment. 
For this question, let us consider only the special relativistic part of the time difference, and ignore the general relativistic part, although the actual measured time difference was the sum of both. 
For example, https://en.wikipedia.org/wiki/Hafele%E2%80%93Keating_experiment gives results only of time gained by the flying clock. It gives no results for time gained by the lab clock, which makes me wonder whether the time dilation was mutual or not.
 A: Special relativity does not say that all motion is relative. It says that all inertial motion is relative. None of the clocks in the Hafele-Keating experiment were moving inertially. (The clocks that remained in the lab at the US Naval Observatory were accelerating due to the earth's rotation. The clock that flew to the west actually had a lower proper acceleration than the lab.)
I think all of this becomes much more transparent if you talk in terms of the metric. The elapsed time on a clock is given by $s=\int_A^B \sqrt{dt^2-dx^2-dy^2-dz^2} $, where $(t,x,y,z)$ are the Minkowski coordinates, and the integration is along the clock's world-line. This integral does not come out the same if you change the path taken from A to B. It is maximized for an inertial path from A to B.
A: Let's ignore the fact that the Earth is rotating, for simplicity.
At the start, the airplane and the airfield are in the same inertial reference frame.  The airplane then accelerates and keeps accelerating very slightly as it performs a large circle.  It is constantly changing reference frames.  In every frame the airplane is in, the airbase is aging more slowly, but the airplane doesn't remain in any frame, and the changes in frame mean that the airbase is a touch older with each new frame.
Eventually, the airplane comes back to the airfield and lands and comes to a stop, again accelerating.  Everything's again in the same reference frame, but the airfield has stayed in that frame.  The airplane has accelerated through its flight, changing reference frames.  Therefore, we're checking the clock that has not accelerated against the clock that has accelerated.
If the plane activated its space drive and headed away from the airfield at a constant speed and direction, then the plane and airfield would each see the other as aging more slowly, but they'd never meet up again to compare clocks.  If the airplane turned around, in that turn it would change reference frames, and that change would show the airfield as aging faster.
