Let $X$ be the set of odd positive fractions with denominator $3$ in lowest terms, except for $\frac13$.
$X=\{x\in\Bbb Q^+:3x\in3\Bbb N\pm1\}\setminus\{\frac13\}$
If $a,b$ are 2-adic units, then any 2-adic number $x$ can be written in the form $$\sum_{i=0}^\infty b_i (2a)^i$$
where $b_i$ is either $b$ or $0$.
Let $T_{a,b}(x)$ be the 2-adic isometry which sends this $x$ to:
$$\sum_{i=0}^\infty c_i 2^i$$
where $c_i$ is either $1$ when $b_i=b$, and otherwise remains $0$.
These functions $T_{a,b}(x)$, with the group operation composition, and $a,b$ still ranging over $\Bbb Z_2^\times$, are the isometry group you get generated by changing uniformizer and coefficients from $2a,b$ to $2,1$ while keeping the underlying binary string the same. The group operation of composition is achieved via multiplication of the two coefficients and the two 2-adic units underlying the uniformizers independently.
Question
Show that $T_{3,1}(x)$ is a topological embedding of $X$ in itself. i.e. $T(X)\subset X$.
My Attempt
I think anybody attempting this question will have more powerful tools in their kit bag than I, but I have a handful of potentially useful tidbits.
I've arranged the problem so the numbers in the domain and the image of $T$ share the same terminal string or periodic part. Intuition said that was the way to go.
The proof will possibly hang around looking at a 2-adic number in $X$ as being of the form $-\frac23\cdot2^t+n:t\in\Bbb N, n\in\Bbb N$. Notice that to get a unique "factorisation" into $n,t$ we would need to pick the "end" of the string $\overline{01}_2$ to the left. This is done by realising there are two cases - the string $\overline{01}_2$ must be followed by a one, or by two zeroes. The $\overline{01}00_2$ case is not of interest as these are negative rationals and therefore not in $X$. So we're looking at $\overline{01}1_2=\frac13$ and need to show that every value in $X$ is cauchy in $X$ and cannot "cross" the barrier $\frac13\cdot2^t$ under the action of $T$.
Remember the isometry property is in play. This means if we find corresponding pair $x,y$ such that $T(x)=y$ then there is no other $x$ in the domain and there is no other $y$ in the image.
On reflection, even the weaker theorem, that $x$ and $T(x)$ differ by an integer is likely to be sufficient.
In sharp contradiction, I claim that $T(X)$ and $X$ are disjoint. (It doesn't matter if $1/3$ is included -- I don't understand the reason it was excluded. I suspect there were some miscalculations by OP.)
Every $2$-adic integer has a unique representation of the form $\sum_{n=0}^\infty b_n 2^n$ with each $b_n \in \{0,1\}$ and also a representation of the form $\sum_{n=0}^\infty b_n 6^n$ with each $b_n \in \{0,1\}$. I will refer to the former representation as the base-2 expansion and the latter as the base-6 expansion. The map $T:\mathbf Z_2 \to \mathbf Z_2$ is defined by $T(\sum_n b_n 6^n) = \sum_n b_n 2^n$.
The fact that $T(X) \cap X = \emptyset$ follows from the following two claims.
Claim 1: Each element of $X$ (indeed any rational number) has a repeating base-2 expansion.
Claim 2: Each element of $X$ has a non-repeating base-6 expansion.
The proof of Claim 1 is completely standard. Here is one way of proving it. Let $q$ be a rational. Then $-q$ has a repeating base-2 expansion in $\mathbf R$. This implies that we have a represenation of $-q$ of the form $$-q = \alpha + \beta (1 + 1/2^k + 1/2^{2k} + \cdots) = \alpha + \beta / (1-1/2^k)$$ with $\alpha \in \mathbf Z[1/2]$ (i.e., $\alpha$ is a rational with a finite base-2 expansion) and $\beta$ is an integer with $0 \le \beta < 2^k$. Now this implies that, 2-adically, $$q = -\alpha + \beta 2^k / (1-2^k) = -\alpha + \beta (2^k + 2^{2k} + 2^{3k} +\cdots),$$ which shows that $q$ has a repeating base-2 expansion in $\mathbf Z_2$.
The proof of Claim 2 is similar. Let $n$ be an integer and suppose $n/3$ has a repeating 2-adic base-6 expansion. Then we have a representation of the form $$n/3 = \alpha + \beta (1 + 6^k + 6^{2k} + \cdots) = \alpha + \beta / (1-6^k),$$ where $\alpha,\beta \in \mathbf Z$. But the right-hand side is a rational number with $3$-adic valuation $\ge 0$, so $n$ must be divisible by $3$.