I think I might have figured it out. Would someone let me know if this proof suffices?

We know for any positive integer . Since , which is the inverse limit space, the division algorithm can be extended as follows. If the 2-adic expansion for , and I want , using the division algorithm, I would have

Since is a division ring, we know there exists such that . Plugging in for and multiplying both sides by , we get:

(Equation 1)

Since and are all integers, it must be that is an integer. Solving for , we get:

for any and .

Also,

So, we have . Next, if we divide both sides of (Equation 1) by , we get:

Since and , by the 3-adic extended division algorithm, we know that . So, from all of this, we obtain that . So, since , the multiplication by a unit of the local ring is essentially rotating the digits of the repeated pattern places to the right, which is exactly what I was trying to prove.

Edit: I am using the notation that is the space of 2-adic integers, not the set of congruence classes of , which is .