Suppose I have two lists $\alpha\_1,\dots,\alpha\_k$ and $\beta\_1,\dots,\beta\_k$ of real numbers such that all $2k$ numbers are mutually algebraically-independent over the rationals. For each $i \in \{1,\dots,k\}$, let $\phi\_k$ be the affine linear map $x \mapsto \alpha\_i x + \beta\_i$.


Let $I$ be the image of the integers, $\mathbb Z$, under some finite sequence of applications of these maps and/or their inverses. (That is, $I = \phi\_{i\_\ell}^{e\_\ell} \circ \cdots \circ \phi\_{i\_1}^{e\_1} ({\mathbb Z})$ for some $\ell \geq 1$, $e\_j \in \{+1,-1\}$, and $i\_j \in \{1,\dots,k\}$).


Is the sequence of maps $\phi\_{i\_1}^{e\_1}, \dots, \phi\_{i\_\ell}^{e\_\ell}$ is reconstructible from $I$? (Is there a unique sequence of applications of these maps and inverses such to find $I$ from ${\mathbb Z}$?)


It seems that a formal statement would be that these maps generate a free group within the group of affine linear maps. My intuition is that two distinct ways to generate an element of $I$ with these maps would create an algebraic relation between these algebraically-independent elements, but explicitly building the relation is cumbersome.



Algebraically Independent Numbers and Affine Linear Maps
The sequence of maps $\phi\_{i\_1}^{e\_1},\dots,\phi\_{i\_{\ell}}^{e\_{\ell}}$ is not uniquely determined by the image set $\phi\_{i\_{\ell}}^{e\_{\ell}}\circ\dots\circ\phi\_{i\_1}^{e\_1}(\mathbb{Z})$. One reason for this is that all commutators $\phi\_i\phi\_j\phi\_i^{-1}\phi\_j^{-1}$ are translations $x\mapsto x+b$, so any two commutators commute with one another. Thus, for example,
$$
\phi\_1\phi\_2\phi\_1^{-1}\phi\_2^{-1}\phi\_3\phi\_4\phi\_3^{-1}\phi\_4^{-1} =
\phi\_3\phi\_4\phi\_3^{-1}\phi\_4^{-1}\phi\_1\phi\_2\phi\_1^{-1}\phi\_2^{-1}.
$$


I would guess that commutativity of commutators is the only source of counterexamples to the proposed question.


