I don't think there's a way to meaningfully connect your comment to general relativity (GR). "The answer" in GR is that Mercury is in geodesic motion ("free fall"), with its geodesic picked out by the mass of the solar system and principally its central mass (the sun). For all practical purposes, nothing about Mercury itself selects which geodesic its on.

The weak equivalence principle (WEP) says that in curved spacetime a small freely falling object's orbit around a large central pointlike mass is completely determined by the former's initial position in spacetime and its initial velocity.

The strong equivalence principle (SEP) says that this remains true for the smaller body even if that body is bound by its own self-gravitation: the orbit is determined fully by the initial position and velocity and not by the small body's internal composition.

We can of course replace the large-mass pointlike generator of a (exterior) Schwarzschild-like spacetime with an extended body that generates some perturbation of a more general central-mass spacetime (like Kerr or Kerr-Newman).

The WEP has been tested extensively in terrestrial labs (torsion balances) and satellites in Earth orbit (e.g. MICROSCOPE).

The SEP has been tested extensively using satellite and lunar laser ranging, and is supported by astrophysical observations including the triple-relativistic-star system comprising an inner white dwarf and PSR J0337+1715 orbited by an outer white dwarf.

That the SEP holds up so well means that we can replace Mercury with any mass much smaller than the sun (it does not have to be as small or smaller than Mercury) at any density and with any internal configuration as long as it is self-bound gravitationally and/or electromagnetically. So we could replace Mercury with a small black hole or a "hot Jupiter" and the orbit would be identical.

The sun generates a perturbed Kerr metric that must take into account not just its rotation but the sun's non-sphericity (the "solar bulge" makes it slightly oblate). Given the metric (or a good enough approximation) we can solve the geodesic equation, and see that Mercury's orbit follows one of the generated geodesics to high precision.

In <http://dx.doi.org/10.1103/PhysRevLett.120.191101> Clifford Will writes: "Finally, at a purely pedagogical level, it is often stated that the relativistic perihelion advance of Mercury is really only a test of the vacuum Schwarzschild solution (or of the slow rotation limit of the vacuum Kerr solution, if one wishes to include the frame-dragging effect), since all the relativistic effects can be derived simply from those metrics." Note that Will's paper uses post-Newtonian corrections to the 2PN level, which is perfectly reasonable given how small Mercury's v/c^2 is, and the ease with which the approach deals with perturbations from the other planets. See also Will's 1986 book, chapter 5 of which is devoted to Mercury's orbit.

> Mercury is heavier when it is at perihelion ... because it's moving faster

"Moving faster" is not a frame-independent statement. We can always use a freely falling coordinate system where Mercury is always at the origin, or a freely-falling coordinate system where in the neighbourhood of a point Mercury experiences no acceleration against those coordinates. There are of course an infinite number of systems of coordinates in which Mercury, the Sun, or both accelerate(s) against that set of coordinates over the course of an Earth year. The point of relativity is that physics do not depend on a choice of coordinates.

"Heavier" is at best ambiguous, requires a lot of care in stating it covariantly (compare the stress-energy tensor), and is in any event irrelevant if the Strong Equivalence Principle holds. More technically, the backreaction of Mercury on the metric generated by the sun (or sun + other planets) is negligible.

If you explain what you mean by Mercury's "relativistic mass" and roughly the magnitude you think its change should be through Mercury's orbit, someone might be help clear up what is probably a misconception. Bear in mind that Mercury moves very slowly compared to c.

Finally, with respect to your last paragraph:

Mercury orbital eccentricity: 0.205630. Pluto orbital eccentricity: 0.2488. If Earth has no perihelion precession because of the low eccentricity (0.01671) of Eearth's orbit, and Mercury's perihelion precession is driven by its higher eccentricity, what do you think Pluto's perihelion precession should be: higher or lower than Mercury's? And why?