= Solution
Set $f(r)=1-2M/r$ and use <proper time> along a timelike <geodesic>. In the equatorial plane, the radial <Christoffel symbols> needed for a circular orbit are
$$
\Gamma^r_{tt}=\frac12ff'=\frac{fM}{r^2},\qquad\Gamma^r_{\varphi\varphi}=-fr.
$$
The radial <geodesic equation> with $\dot r=0$ gives $(fM/r^2)\dot t^2-fr\dot\varphi^2=0$. Outside the horizon this implies
$$
\boxed{\left(\frac{d\varphi}{dt}\right)^2=\frac{M}{r^3}.}
$$
For stability use the conserved energy $E=f\dot t$ and <angular momentum> $L=r^2\dot\varphi$. The normalization of the timelike tangent in the <Schwarzschild metric> gives $\dot r^2+V(r)=E^2$, with the <effective potential for timelike Schwarzschild geodesics>
$$
V(r)=f(r)\left(1+\frac{L^2}{r^2}\right).
$$
At a circular orbit $V'=0$, so $L^2=Mr^2/(r-3M)$, requiring $r>3M$. Differentiate $V$ holding this conserved $L$ fixed and then substitute the circular value:
$$
V''(r)=\frac{2M(r-6M)}{r^3(r-3M)}.
$$
The linear radial perturbation equation is $\delta\ddot r=-V''(r)\delta r/2$. Thus \b[circular timelike orbits are stable for $r>6M$], unstable for $3M<r<6M$, with the marginal case at $6M$.
For the orbiting astronaut at radius $R$, use $d\varphi/dt=\sqrt{M/R^3}$ to obtain
$$
\frac{d\tau_a}{dt}=\sqrt{1-\frac{2M}{R}-R^2\left(\frac{d\varphi}{dt}\right)^2}=\sqrt{1-\frac{3M}{R}}.
$$
The static Earth twin has $d\tau_0/dt=\sqrt{1-2M/R_0}$. During equal coordinate-time intervals of the long circular-orbit phase, the astronaut ages less precisely when
$$
1-\frac{3M}{R}<1-\frac{2M}{R_0}\quad\Longleftrightarrow\quad\boxed{2R<3R_0.}
$$
The outward and return transfers have finite additional proper times; the stated long-orbit comparison is the limit in which the circular phase dominates them. At equality the circular phase contributes no age difference, so transfer details can decide the finite-trip result.
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