Can civilizations orbiting black holes keep time with their interstellar travellers? Due to relativistic-velocity interstellar travel time dilation, travellers would return in the far future of any civilization launching such an effort. A chart from wikipedia indicates typical time dilation ratios for various destinations:

Could an advanced civilization orbiting a black hole position themselves within its gravity well such that they enjoy constant time with their travellers?
 A: If we assume everything takes place in the Schwarzschild metric, then the time dilation factor is obtained from the metric as
$$c^2 d\tau^2 = c^2\left(1 - \frac{r_s}{r}\right)^2dt^2 - \left( 1-\frac{r_s}{r}\right)^{-1} dr^2 -r^2 d\theta^2 - r^2\sin^2\theta d\phi^2\ ,$$
where $d\tau$ is  proper time interval and $r_s$ is the Schwarzschild radius.
To answer your question we need to make some simplifications. Let's assume the "stay at home" observer is in a circular orbit around the black hole at fixed $r$. In which case for them, using $v_{\rm orb} = rd\phi/dt$ and $dr=d\theta=0$, we obtain
$$ d\tau = \left( 1 - \frac{r_s}{r} - \frac{v_{\rm orb}^2}{c^2}\right)^{1/2}dt = \left(1 - \frac{3r_s}{2r}\right)^{1/2}dt\ .  $$
For the body racing away and then coming back, I'm going to assume a uniform speed $v$ on a radial trajectory. Of course you can make it more complicated by having say uniform acceleration or soemthing, but that doesn't change the principle of the matter. So here we have $d\theta=d\phi=0$, and because the vast majority of any flight will take place with $r\gg r_s$, then we can just assume the flat spacetime metric and get
$$ d\tau = \left(1 - \frac{v^2}{c^2}\right)^{1/2} dt \  $$
for the proper time interval of the travelling observer.
In order for the proper time interval to be the same for both observers, then we need
$$  1 - \frac{3r_s}{2r} \simeq 1 - \frac{v^2}{c^2}\ , $$
$$ r = \frac{3r_sc^2}{2v^2}\ .$$
In the Schwarzschild metric, the innermost stable circular orbit is at $r=3r_s$. So long as $v < c/\sqrt{2}$ then a solution could be found.
Of course, in your example, speeds of $c/\sqrt{2}$ are actually reached pretty quickly if you accelerate constantly at $1g$. If you want to match more extreme time dilations (any more than a factor of $\sqrt{2}$ in fact), then you would need to have your stay-at-home observer either somehow kept stationary at a a radius not much larger than $r_s$ or in a close orbit around a Kerr black hole for which orbital stable radii could be much smaller, but would require an almost maximally spinning black hole to get very large time dilation factors.
In order to avoid nonsensically large tidal forces, this will also have to be a supermassive black hole.
