Let $\sigma$ be a permutation of $\mathbf Q.$ We call $\sigma$ bounded (the term might be somewhat misleading, but however it is used in a couple of papers) if there is a real number $M$ such that $ \mathrm{dist}(\sigma x,x) =|\sigma x-x| \le M $ for all $x \in \mathbf Q.$
Now can a bounded infinite cycle $ \pi=(\ldots,-2,-1,0,1,2,\ldots) $ be written as a product of bounded involutions (permutations of order two)? Notice that "nice" bounded products of, say two, disjoint infinite cycles can be written as products of bounded involutions, e.g. the permutation $ (\ldots,-4,-2,0,2,4,\ldots) (\ldots,-3,-1,1,3,\ldots). $
EDIT: to clarify a bit chandok's argument below: suppose $\pi$ is a product of two involutions $\sigma$ and $\tau$: $\pi=\sigma \tau.$ Then $ \sigma \tau \sigma(n+1)=\sigma(n+1)+1. $ for all integers $n.$ On the other hand, $n+1=\pi(n)$ and $\tau \sigma=\pi^{-1},$ whence $ \sigma (\tau \sigma) \pi(n) = \sigma \pi^{-1} \pi(n)=\sigma(n)=\sigma(n+1)+1. $ Thus $\sigma(n+1)=\sigma(n)-1$ and $\sigma(n)=a-n$ for all naturals $n.$ Thus $\sigma$ is unbounded.
The argument seems to be not easily adaptable for products of more than two involutions.