Let $f_{a,b}(x)=ax-b\cdot2^{\nu_2(x)}$ act on the 2-adic numbers.
Then I have as a fairly resolute theorem that for any choice of two 2-adic units $a,b$, every $f_{a,b}$ is topologically conjugate to the others. I.e. there is a homeomorphism $T$ that topologically conjugates any given $f_{a,b}(x)$ to another.
To give a little insight into the function, it commutes with $2x$ in the sense that $f(2x)=f(x)$ or in other words it acts faithfully on the obvious quotient $\Bbb Q_2\to\Bbb Z_2^\times$ and therefore its action on the quotient $\Bbb Q_2/{\langle2\rangle}$ is sufficient to define it.
It follows that these functions generate a subgroup of the automorphism group of $\Bbb Q_2/\langle2\rangle$. If we let $T_{1,1}(x)$ be the identity homeomorphism which conjugates $f_{a,b}$ to $f_{a,b}$ then each homeomorphism is written:
$T_{a,b}(x)=\displaystyle\sum_{n=0}^\infty2^{\nu_2(f_{a,b}^n(x))}$
Where $n$ indicates compositions of $f$.
I'd like to understand the group these generate.
Elements of the form $T_{1,b}$ are straightforward because $T_{1,b}=b\times T_{1,1}$ and therefore the inverse is $T_{1.\frac1b}$. Which is helpful because this category of operations is therefore also commutative and closed to $b\in\Bbb Z_2^\times$.
But inverses of $T_{a,1}$ and $T_{a,b}$ I'm not so sure about, nor commutativity and closure of say $T_{a_1,1}\circ T_{a_2,1}$.
Question
What's the automorphism group of $ax-b\cdot2^{\nu_2(x)}: a,b\in\Bbb Z_2^\times$ on $\Bbb Q_2$? In particular, do the generators $T_{a,b}:a,b\in\Bbb Z_2^\times$ generate elements beyond themselves and if so, what is the set and the operation?
Edit
I attempted an answer based upon composing two affine transformations $ax+b$ and $cx+d$ but I was able to confirm this fails:
If $T_{a,b}$ topologically conjugates $f_{a,b}(x)=ax-b\cdot2^{\nu_2(x)}$ to $f_{1,1}(x)=x-2^{\nu_2(x)}$ then for composition do we we have:
$T_{a,b}\circ T_{c,d}=T_{ca,cb+d}$ ?
This would generate the group having the set $T_{a,b}:a\in\Bbb Z_2^\times, b\in\Bbb Z_2$
Note that $T_{1,1}(x)$ is the identity function because it sends the standard binary representation of any 2-adic number to itself.
And NO, this doesn't work because it fails to agree with $f_{1,b}\circ f_{1,d}=f_{1,b\times d}$
Curiously, this attempt yields addition of $b,d$ instead of multiplication.