An Interesting Detail

Let's consider what that would mean in the general case. If for some reason the reflections of an expansion are equivalent to the rotations, then we can take any reflection and any rotation and make them identical by removing finite prefixes. Then, however, by removing up to one iteration more we can show that the primary reflection is equal to some rotation. Then we can just write down the coefficients, mess with the indices a bit, and show that there are three possibilities. If the expansion length is even, then there are either two or zero rotations that are equal to their own reflections.

[1,2,3,4,4,3,2,1]
[4,3,2,1,1,2,3,4]
 
[1,2,3,4,5,4,3,2]

Or, if the expansion length is odd, then there's always exactly one rotation that's equal to its own reflection.

[1,2,3,4,3,2,1]

It can also happen that the primary reflection is equal to more than one rotation. In that case, the three possibilities apply to each rotation. Also, since the rotations are equal to the same thing, they're equal to one another, so the expansion has rotation symmetry.

How does all that apply to expansion(u) (with u ≥ 5)? Well, there can't be any rotation symmetry because the number of large units (22) and the number of small units (11) are relatively prime. (See the point in Symmetries about the number maze.) The expansion length is even, not odd, and the “zero rotations” case is unreachable because those expansions aren't composed of pairs. The only possibility left is that there are exactly two rotations that are equal to their own reflections. In other words, we've just proved that the third symmetry holds! The proof has the disadvantage of depending on two unknowns, but it also has the advantage of providing a tangible reason that the symmetry has to hold.

For comparison, I do still think it should be possible to use half-expansions to construct a proof. That proof wouldn't depend on any unknowns, but it would also be just a bunch of cases and not provide any reasons.