You said (substituting a set of twins for objects).
"The moving twin Abe is younger in the frame of the twin Bill at rest, not in his own frame. In his own frame he ages at the normal rate."
The twins age at their normal rate regardless. They are always in their own frame. Relativity is about making an observation and there is no reason why the moving twin couldn't make the observation that he is aging slower than his brother. All the twins need is some knowledge of relativity and some information.
Abe signs on with the crew of 'Warp Drive 1' while Bill remains on Earth. Warp Drive 1's maiden voyage is to visit a solar mass (M = 1477 meters) black hole, free fall to a knife edge orbit just outside the photon sphere at r = 3.000001M, remain in orbit for 172,800 seconds (Two Earth days) wristwatch time (dTau), and then return to Earth arriving ~ 9.5 years in Earths future. During the warp phase of the journey the ship remains in an inertial rest frame so the difference in wristwatch rate for Abe and Bill is minimal during this phase of the journey.
Lets make a prediction (observation) before the ship even leaves to test General Relativity. What will be the difference in elapsed wristwatch time, dTau, for Abe with respect to the elapsed coordinate time, dt, for Bill when they meet upon Abe's return? Keeping in mind that dTau is measured with Abe's wristwatch (shipframe) and the coordinate time for Bill is measured with Bill's wristwatch (Earthframe).
From the Schwarzchild geometry
dt/dt = (1-3M/r)^1/2
Substituting 172,800 seconds for dt and 3.000001M for r and solving for dt
dt = 172,800s / 5.7735x10^-4 = 299298529.2s
Since there are 3.156x10^7 seconds / Earth year
299298529.2 seconds / 3.156x10^7 seconds / Earth year ~ 9.48 Earth year
When Abe returns the twins can confirm the prediction (observation) by just comparing clocks.