Is $Q(x)$ an isometry on $\Bbb Q_2$?

Question

Wikipedia defines $Q(x)$ this particular isometry on $\Bbb Z_2$ equivalently to the following:

Let $T(x)=\cases{\frac{3x+1}2 & if $x\equiv1\pmod2$ \\\frac x2 & if $2^{\nu_2(x)}>1$\\2x&if $2^{\nu_2(x)}<1$}$

be a continuous and measure-preserving function on the ring of 2-adic integers.

Then let $Q(x)=\sum_{n=0}^\infty T^n(x)\pmod2\cdot2^n$ which makes $Q(x)$ an isometry on $\Bbb Z_2$.

Problem

Does $Q(x)$ extend to $\Bbb Q_2$?

Attempt

I think there's an obvious approach which begins multiplying $x$ by the smallest $2^k$ such that $2^kx\equiv1\pmod2$. Then proceeds as before, then gives the result $2^{-k}Q(x)$ - is this flawed?

So in order to write $Q$ as the sum piecewise, while following the function $T$, as before, I think we want:

Let $Q(x)=\sum_{n=-k}^\infty 2^{k}T^n(x)\pmod{2}\cdot2^{n-k}$

Is this now an isometry from $\Bbb Q_2\to\Bbb Q_2$?

Answer 1

Yes, I have since determined $Q$ is an isometry on $\Bbb Q_2$ and a good, succinct way of writing it is:

$$Q(x)=\sum_{n=0}^\infty 2^{\nu_2(c^n(x))}$$

Where $c^n(x)$ indicates the $n^{th}$ composition of $c(x)=3x+2^{\nu_2(x)}$, and where $2^{\nu_2(x)}$ is the highest power of $2$ that divides $x$.