he keyhole implies that both rims have an elevation which matches exactly with the current line of sight; it could be in theory, but keep such *vertical* alignment for sols and sols is something I can't explain.
Actually, it just requires that the top of the far rim be below the maximum angle of inclination for the front rim, and above (on average) the minimum angle of inclination for the near rim, after correcting for the curvature of Mars. See the enclosed figure:
From a quick back-of-the-envelope calculation, the far rim will be reduced in apparent height by a distance,
h = (rfar^2-rnear^2)/(2*R)
where rnear = distance to near rim, rfar = distance to far rim, and where R = 3.390e6 meters is the radius of Mars. Given that the diameter of the crater is 750m, this gives
h = 0.5 m when rnear = 2000m
h = 0.3 m when rnear = 1000m
h = 0.2 m when rnear = 500m.
Since the relief of the crater is expected to be pretty small, probably only a few meters at maximum, the curvature effect should be pretty important, and unless there were a significant height difference for the far rim, it would never broach the top of the near rim.
Perceptually, I know, it seems especially from early images that the beacon was above the height of the rim. However, I think this is just a result of bleed-over from the highly saturated pixels containing the beacon. The nearer shots, like the one I included above, seem to suggest that the beacon height is roughly the same as the height of the Victoria Crater profile. The pixel resolution is so poor (esp. for those of us using JPEGs) that I don't think you can say much more than that.
Beyond the fact that we can see the beacon, really it's so poorly resolved that it would be hard to argue that we can really say whether it stays at exactly one height or another. The only condition for viewing the far wall would be the height of the far rim be tall enough (including the curvature correction) to peak above the dip in the near rim.