This is incorrect. The analemma would still be a figure-8 even with a circular orbit, but it would be symmetric -- the orbital eccentricity only adds the asymmetry we see in Earth's analemma.
This is because the sun on an axial-tilted plane 'lags behind' then 'catches up' to the ideal 0-tilt sun over the course of the year. At the equinoxes, the sun's motion has a significant vertical component. Therefore, it's horizontal speed is slower than an untilted sun (both suns still travel through the sky at the same speed of 360/365 degrees per day), so it will lose ground and drift back. At the solstices, the sun moves horizontally, but at an higher latitude (equal to Earth's axial tilt) on the celestial sphere, covering more degrees of longitude for the same speed than the ideal sun moving along the equator, hence making up the lost ground. This variation in horizontal speed throughout the year creates the figure-8.
I'm having a difficult time understanding this. I trust that you're knowledgeable about the topic given your comment history, but --
You've argued that the Sun's rate of motion across the sky should have a constant magnitude (under the assumption of an Earth analog with a perfectly circular orbit but still 23.5 degree obliquity). It's not clear to me why this should be the case; on the celestial sphere, I would have an easier time accepting that the Sun should have a constant "horizontal" (i.e., in the sense of right ascension) motion due to the Earth's constant orbital speed, and that the Sun's "vertical" (i.e. declination) motion should be purely sinusoidal: greatest at the equinoxes, zero at the solstices. Have I misunderstood something here?
As an extreme case, imagine a tidally locked planet on a circular orbit of 1 AU with 23.5 degree obliquity. An observer sitting on the equator would see the sun moving on a line overhead, crossing the zenith back and forth, with no figure 8 -- right?
It is quite difficult to visualize and it took me a long time to "get it".
The path of the sun's annual motion relative to the stars is determined solely by my physical progress in orbit around the sun. A planet's axial tilt only changes the 'direction' I'm looking in and thus the reference point of my celestial coordinate system. The sun's path will always be a great circle on that celestial sphere (and always the same relative to the fixed background stars) regardless of which reference frame I choose. I think this is enough to surmise that the sun's angular speed on the celestial sphere is constant regardless of axial tilt (assuming a perfectly circular orbit).
Taken to an extreme, imagine a planet with 90° tilt -- the sun would move vertically and pass directly over the pole, making a constant 'horizontal' motion literally impossible.
I'm not sure what your tidally locked example is meant to demonstrate, since that's literally what the analemma is -- the path the sun would make in the sky once you subtract a planet's local axial rotation, i.e., tidally locking it to the orbital parent.
Thank you for the further explanation - this has convinced me that it's plausible, though my mind is slow and I still need to think about it to be fully convinced.
This is because the sun on an axial-tilted plane 'lags behind' then 'catches up' to the ideal 0-tilt sun over the course of the year. At the equinoxes, the sun's motion has a significant vertical component. Therefore, it's horizontal speed is slower than an untilted sun (both suns still travel through the sky at the same speed of 360/365 degrees per day), so it will lose ground and drift back. At the solstices, the sun moves horizontally, but at an higher latitude (equal to Earth's axial tilt) on the celestial sphere, covering more degrees of longitude for the same speed than the ideal sun moving along the equator, hence making up the lost ground. This variation in horizontal speed throughout the year creates the figure-8.