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It is an old result (from "On one-relator groups having elements of finite order" by J. Fischer, A. Karrass and D. Solitar (1972)), but if $G=\langle X; R^n\rangle$, $n>1$, then the derived subgroup is an infinite free product of cyclic groups of order $n$. Also, $G$ is hyperbolic (and has one end).
Hmm - I really didn't pick up on this! (But it certainly makes sense - thanks.) Sageev talks about "essential" actions (which corresponds to not messing up the hyperplanes, and in the finite dimensional case this corresponds to an action with an unbounded orbit) - was this dropped somewhere along the way?
@HW: The sort of answer I would hope for would be "If you put these restrictions on $G$ and $K$ (and perhaps on the isomorphism between $K$ and $K^{\prime}$) then you can say something." Along the lines of the Baumslag-Tretfoff conditions for residual finiteness; sufficient, but not necessary.
In general, you can use the reidemeister-schreier algorithm to compute the presentation of a subgroup of a given subgroup $H\leq G$. It works by looking at the cosets $G/H$, so it is nicest if $H\lhd G$. If $H$ has finite index in $G$ then it will spit out a finite presentation for $H$. I found a worked example of this here, math.stackexchange.com/questions/59273/…
So...groups with property FA are the only groups which have the property in my question? (I.e. having property FA is equivalent to the property I am wondering about?)