Using relative velocity between two frames and the time elapsed between events in one frame to find the time elapsed/event separation in the other Q from text: A second transporternaut is beamed to a much more remote galaxy that is moving away form Earth at .87c. This time, too, she stays in the remote galaxy for one year as measured by clocks moving with the galaxy before returning to Earth by Transporter. How much has the transporternaut aged when she arrives again back at Earth?
In this question, of which this above is part e, the transporters send the person as data at light speed form place to place, and disassembly and reassembly time is negligible. From a previous part, I found that she will not age on the trip there or back, as she is travelling at the speed of light, and the interval for her is 0 (her time elapsed as seen form earth is, say, 2E6 ly to get to andromeda, but andromeda is 2E6 ly away, so I = (t^2-d^2)^(0.5) = 0). 
All her aging, then, is on this galaxy, for one year galaxy time. I want to find out how long that is in Earth time. In the galaxy's frame, the event of her arriving and leaving occur at the same place, and since one year elapsed the interval is 1 ly light travel time. This must be the same as calculated form Earth. I was trying to find the distance, as seen form Earth, that the galaxy moved. This would be the separation between the events of her arrival and departure form the galaxy, as seen form Earth, and with this and the interval I could find the time elapsed between the events as seen form Earth. 
I could use the relative speed between Earth and this galaxy, to find the distance between them. But what frame will I be finding this distance in? Example, If I use the time elapsed as 1 year, I get .87 ly distance, but this is as seen from this new galaxy. From Earth it wasn't a year elapsed, which is the point to begin with. If observers both on earth and on the galaxy see the same relative velocity between them (I assume they do), but they can experience different amounts of time being elapsed from the arrival and departure of the transporternaut, they must find different distances moved by the galaxy in the meantime. So I shouldn't be able to use the relative velocity, which is the same measured in either frame, and the time as measured in the galaxy, which differs from that measured on Earth, to find the distance as seen form Earth, should I? If I can, how comes? What other info do I have that could solve this problem? 
It seems like I have both the time elapsed and the distance the Earth moved away, as seen form the galaxy, whereas I ahve none of this info for the observer in Earth's frame. 
If I use the .87 ly as the distance between events, seen on Earth, and the interval as 1 ly, I get 1.3 ly time elapsed, so 1.3 years aging. 
 A: You're going to need to more precisely define "aged."
If you mean "how much time has elapsed for her?", which I guess we could phrase as "how much have her cells aged?" then because she doesn't notice time during the "transporter" process (as she has been atomized into constant data) you can just use the local clocks. If she's off in the same reference frame as this other galaxy and its clocks say that she's been there for a year, then that's how much her cells are going to age. There may be some nuance if she has to locally accelerate to the same speed as the galaxy once she gets over there, but it seems like you could probably drive that contribution to be much smaller than a year.
But on Earth, her "age" is defined as the number of times the Earth has orbited the Sun since her birth. This includes the, say, 2e6 years that it takes to travel there, as well as the time dilation effect in the remote galaxy, as well as the return trip.
A strange hybrid of these two definitions -- using the time-dilated time but not the propagation delays -- does not seem warranted unless the question specifically asks for her experienced time, given that she experiences 1 year as measured by clocks moving with Earth. That is, you have to imagine that they send her to RemoteGalaxy, then some Earth time $t$ later send a big bright message of "you're done! come back home!". There are two possibilities for $t$ here: either Earth sends the message after $t = \text{1 year}$ or else Earth tries to satisfy $c (\text{1 year} - t) = v (\text{1 year})$ so that Earth coordinates think that she has received the message after 1 year of Earth time, so they send it after 0.13 years.  Either way, she receives the message at an Earth time T given by $c (T - t) = v T$ or $T = t / (1 - \frac vc)$. However, Earth sees her clocks as ticking slowly by a factor $1/\sqrt{1 - (v/c)^2} \approx 2$, so she "ages" (in the cellular sense above) about half as much during that time, as the contents of her space ship and the remote galaxy appear to be moving "in slow motion" from the Earth perspective.
