Let $f:X\to X$ be an isometry on the 2-adic metric space $X,d$.
Let the periodicity of $x$ be the number of repeating terms in its 2-adic representation.
What rules govern the periodicities of $f(x)$? Particularly rules that relate the periodicity of $x$ to the periodicity of $f(x)$.
For example:
Let $x,y\in X$ be some pair having periodicity $1$ - i.e. terminating to the left in either $\overline1$ or $\overline0$.
Is it possible for the periodicity of $f(x)$ and $f(y)$ to differ?
Is it possible for the periodicity of $f(x)$ to be lower than the periodicity of $f(0)$
Do all 2-adic units $x$ have the same periodicity in $f(x)$?
Periodicity, if I understand you correctly, is a property of rational p-adic numbers, with irrational p-adic numbers being aperiodic. Probably the simplest isometry we can consider is $f(x) = ux+b$ with $|u|_p=1$ and no restriction on $b$.
Since your questions seem to hinge upon having the isometry map different numbers in a coordinated way, I'll show you how to construct an isometry that fixes one number and moves another number to a number with an arbitrary period of your choice. This is like picking an axis of rotation to leave fixed and rotating one point into another, except we are looking at a kind of non-archimedean version of it.
So let's say we want $f(a) = a$ to remain fixed and $f(b)=c$ to be the moved point, with $c$ arbitrary so long as it's equidistant $|c-a|_p = |b-a|_p$ to maintain the isometry. The key point here is that $b$ and $a$ only have to have the same few starting digits (reading from right to left) as $c$ and $a$ do, but after that you can completely alter $c$ to differ from $b$ to have entirely set of digits to alter the period.
$$f(x) = \frac{f(b)-f(a)}{b-a}(x-a)+ a$$
Just to put it into practice, let's look at your example $a=\bar 0$ and $b=\bar 1$ both have periodicity 1. To maintain the isometry I have to choose $|c- \bar 0|_p = |\bar 1 - \bar 0|_p = 1$ and so I arbitrarily pick $c = \overline{01}$ which has a different periodicity 2. I could just as well have picked $\overline{001}$ or $\overline{11011}$ or something entirely aperiodic such as $...000111001101$ alternating between an ever increasing number of consecutive 1s and 0s but never periodic.
$$f(x) = \frac{\overline{01}}{\bar 1}x$$
Since $\bar 1 = \frac{1}{1-p}$ and $\overline{01} = \frac{1}{1-p^2}$ I'll rewrite the isometry that way just to be a little less opaque.
$$f(x) = \frac{1-p}{1-p^2}x$$