About a collection of linear maps $L=\bigcup_{i=1}^\infty X_i$ I recently asked: About this transformation, and I wanted to follow up with another related question:
Consider a collection of linear maps $L=\bigcup_{i=1}^\infty X_i.$ Where $X_1=\big\{(x,y)\mapsto\big(ax,\frac{y}{a^1}\big)\big\}.$ $X_1$ is called a "squeeze mapping" for parameter $a.$ I don't know of any names for $X_2$ and others. The collection of squeeze mappings forms a one-parameter group isomorphic to the multiplicative group of positive real numbers.

Does the collection of linear maps $L$ form an algebraic structure, such as a $R-$module or vector space?

I think that for each $X_i$ we can collect the mappings to form a one-parameter group. That would mean $L$ is a collection of one-parameter groups. Then I could combine these groups with an operation.

The collection of all $X_1$ forms a one parameter group. Are the collection of $X_i$ all isomorphic to each other?

Related: squeeze mapping group theory.
 A: For the first question: Although each set $X_i$ is closed under composition, the whole collection $L$ is not closed. Therefore, there is no group structure. As $R$-modules are abelian groups, we see that $L$ is not an $R$-module for any ring $R$.
(I am assuming that the operation you are considering is composition. However, $L$ is also not closed under pointwise addition, which is the other obvious operation: if we pointwise add $(x, y)\mapsto (2x, \frac{y}2)$ and $(x, y)\mapsto (2x, \frac{y}4)$ then we get $(x, y)\mapsto (4x, \frac{3y}2)$, which is not in $L$.)

For the second question: Yes, the $X_i$ under composition (and with $a\neq0$) are all isomorphic as groups to $(\mathbb{R}\setminus\{0\}, *)$, and hence isomorphic to each other. The obvious map, the one defined by $\phi_i:a\mapsto \left((x, y)\mapsto(ax, \frac{y}{a^i})\right)$, works. It should be clear that it is an bijection, while is it a homomorphism as:
\begin{align*}
\phi_i(a)\phi_i(b)
&=\left((x, y)\mapsto(ax, \frac{y}{a^i})\right)\circ\left((x, y)\mapsto(bx, \frac{y}{b^i})\right)\\
&=\left((x, y)\mapsto(a(bx), \frac{y/b^i}{a^i})\right)\\
&=\left((x, y)\mapsto((ab)x, \frac{y}{(ab)^i})\right)\\
&=\phi_i(ab)
\end{align*}
