Could the Earth-Moon rotation period equal one Earth year? I just stumbled on this  rather confused QA. The point seems to be if the Earth-Moon rotation period was identical to the Earth-year, then ...

... all three would stay in a line at all times.
Could this actually happen?  (ie: they stay in a line at all times.) It seems "wrong" somehow.
Or do we simply not know ... the solution would only be numeric since it's a three-body problem?
Assuming the solar system was otherwise completely empty, so no other perturbances; assuming the actual masses of the three.
{If the answer is "yes", where the hell is the CG, what figure does it make?}
 A: That configuration is not stable, as the moon would have to be at Lagrangian point $L_1$ or $L_2$, which are not stable. Any perturbation would cause an exaggerated deviation from those orbits, ending either in an earth moon system rotating around each other, or the Earth and moon orbiting the sun separately.
A: This is the phenomena seen in spiral galaxies, where circular speed at different distances from the center of galaxy is same (pretty much). To explain this phenomena, physicists had to come up with dark matter hypothesis. So, may be with bunch of dark matter with right kind of spread, it is theoretically possible. 
Because physicists had to come up with dark matter hypothesis to explain this phenomena, you can safely bet, it is not explainable otherwise.
A: It could work - while it is a three body problem, it is a solvable one.
You can make a hand-waving approximation by just looking at the earth-moon system and figuring out the distance the moon needs to be from the earth to have an orbital period of 1 year. This turns out to be about 0.014 AU - which is large enough to suggest that the sun's gravitational field is enough to perturb things by maybe a few percent between opposite ends of the orbit.
But that's a very messy way of dealing with it. There is a cleaner solution. All three masses (which I'll treat as point masses) are in a line, and all are orbiting the center of mass of the earth-moon-sun system. To simplify things, lets pretend the sun is right at the center of mass; this isn't going to make much difference.
Lets pick the scenario where the earth is closer to the sun than the moon is. In effect then, we have the earth at distance $r_{e}$ and the moon at distance $r_{m}$ orbiting with the same period, and all the forces involved are conveniently completely radial.
The earth is orbiting at $r_e$ with a centripetal force given by the gravitational force from the sun minus that from the moon. The moon is orbiting at $r_m$ with a centripetal force given by the gravitational force from the sun plus the earth.
$m_e\omega^2 r_e=\frac{Gm_s m_e}{{r_e}^{2}}-\frac{Gm_m m_e}{(r_m-r_e)^{2}}$
$m_m\omega^2 r_m=\frac{Gm_s m_m}{{r_m}^{2}}+\frac{Gm_m m_e}{(r_m-r_e)^{2}}$
Since the various masses are known, and $\omega$ is the angular velocity which is the same for both objects (by definition of the problem) and $\omega=\frac{2\pi}{T}$ where $T$ is the orbital period in seconds (1 year, for the sake of argument). Then you need to solve the pair of equations to find $r_e$ and $r_m$.
You can throw in the orbit of the sun around the center of mass of the solar system to give you an exact solution for point particles. It is an nice equilibrium solution, but may or may not be a stable equilibrium.
The real world is more complex of course, with real bodies not being point masses, other bodies existing in the solar system, magnetic fields, tidal effects and other fun things that might slowly make a difference.
