Given an infinite set $X$ consider the set of sequences $X^\mathbb{N}$ and the map $s$ that shifts all elements up by $1$ or rather $(a_i)\mapsto (a_{i+1})$. The problem asks to find a map $s_{1/2}$ such that $s^2_{1/2}=s_{1/2}\circ s_{1/2}=s$. Here is my attempt:
Proof: I use the fact that the equivalence classes of the relation $\sim$ (where $(a_i) \sim (b_1)$ iff there are $n,m\in \mathbb{N}$ such that $s^n(a_i)=s^m(b_i)$) have cardinality $\vert X \vert$. I partition all the equivalence classes of $X^\mathbb{N}$ into ordered pairs. I fix for each pair $(A,B)$ of equivalence classes a bijection $f_{A,B}:A\rightarrow B$. I define $s_{1/2}$ as follows. If $(a_i)$'s equivalence class is the first element of the pair then $s_{1/2}(a_i)=s(f(a_i))$. If instead the equivalence class is in the second part of the pair we set it to be $s_{1/2}(a_i)= f^{-1}(a_i)$. I think it is immediate that $s_{1/2}$ has the desired property.
Observation: An easy modification of the proof can be used to show that $s_{1/n}$ for all $n$. One instead of partitioning the equivalence classes into ordered pair they partition them into ordered $n$-tuples $(A_1,\dots,A_n)$ and fix bijections $f_1:A_1 \rightarrow A_2$, $f_2:A_2\rightarrow A_3\dots f_n:A_n\rightarrow A_1$ where we also require that $f_n=(f_{n-1}\circ \dots \circ f_1)^{-1}$ we define of course $s_{1/n}$ as follows: If $(a_1)$ is in an equivalence class that is not the last element of its $n$-tuple and is an $A_i$ then we set $s_{1/n}=f_i(a_i)$ and if it is the last element of the $n$-tuple then we set $s_{1/n}(a_i)=s(f(a_i))$.
The true problem is there a set of bijections $s_q:X^\mathbb{N}\rightarrow X^\mathbb{N}$ for each $q\in \mathbb{Q}$ but such that $s_1=s$ and for all $a,b\in \mathbb{Q}$ we have $s_a\circ s_b=s_{a+b}$ and for which sets $X$ does such a set exist. I believe all infinite sets have such a set of maps.
Proof: The only thing we used of the map $s$ is that it sends elements of the same equivalence class into the same equivalence class or rather $(a_i)\sim ( b_i)$ then $s(a_i)\sim s(b_i)$ and that it was a bijection. So I define by induction $ s_{1/(n+1)!}$ once I have defined $s_{1/n!}$ and such that $s^{n+1}_{1/(n+1)!}=s_{1/n!}$. This ought to be enough to prove the problem since taking all such maps and their powers covers all rational numbers and they are compatible by construction. To build inductively $ s_{1/(n+1)!}$ from $s_{1/n!}$ we assume that all the equivalence classes are subdivided into $n!$ tuples. We partition all of the $n!$ tuples into $(n+1)$ tuples $(A_i,\dots,A_{n+1})$ where $A_i$ are in this case $n!$-tuples. We can now define a bijections $f_i:A_i\rightarrow A_{i+1}$ and $f_{n+1}=(f_1\circ \dots \circ f_n)^{-1}$ and again define the map $s_{1/(n+1)!}$ to be the map $f_i$ if $(a_i)$ is in one of the equivalence classes in the $A_1,\dots ,A_n$ and if $(a_i)$ is in $A_{n+1}$ we set the map to be $s_{1/n!}(f_{n+1}(a_i))$ which is analogous to the constructions done before.
Is this proof correct and if so are there any more elegant ways of exposing it? Any help with title and tags would be greatly appreciated as well