They must arrive at the same time as they are mirror images of each other in all respects.
They are only mirror images of each other in point C's reference frame. Don't forget, in a given object's reference frame, it is the one standing still and other objects are the ones that are moving. So from ship A's perspective, it is sitting still while point C and ship B are both moving towards it at different speeds. So ship B doesn't look equivalent to ship A in ship A's reference frame.
This is true in all reference frames.
No it isn't. Each ship will see itself arrive at point C before it sees the other ship arrive. Let's say both ships are travelling at 0.9c in the reference frame of point C. In ship A's reference frame, it is sitting still while point C is moving towards it at 0.9c instead. Using the equation (u+v/1+uv/c^2), ship B is seen by ship A to be moving towards it at ~0.994c. Let's also say that the initial distance between each ship and point C is 1 million kilometers. Point C has to travel 1 million kilometers to reach ship A and, at 0.9c, this will take ~3.7 seconds. Ship B, on the other hand, has to travel twice the distance (2 million kilometers) and, at 0.994c, ship B will arrive at ship A's position in ~6.7 seconds. Thus, in the reference frame of ship A, it arrives at point C (or point C arrives at it, depending on how you interpret it), before ship B does. The only way to make both ships arrive at the same time in ship A's reference frame would be if ship B was moving twice as fast as point C (1.8c, which is what you would get from simple velocity addition). Since this is a forbidden speed in relativity, it can't happen and both ships must therefore arrive at different times in each other's reference frames.
and you really did read 147?......and you thought about it?
Again, "I don't see anything in that post about how action-reaction has anything to do with velocity addition at relativistic speeds."