Can a (non-measurable) autonomous flow have a non-trivial periodic orbit without a minimal period? Given a set $X$, a function $x \colon \mathbb{R} \to X$ is periodic if there exists $\tau>0$ such that $x(t+\tau)=x(t)$ for all $t \in \mathbb{R}$; and if $\tau$ is the smallest positive number with this property, we say that $\tau$ is the least period of $x$.
There obviously exist non-constant periodic functions without a least period - for example, taking $X=\{0,1\}$, the function $\mathbf{1}_\mathbb{Q} \colon \mathbb{R} \to \{0,1\}$ is clearly periodic, with every positive rational number being a period.
But now, let us consider "periodic orbits". Given a set $X$, a function $x \colon \mathbb{R} \to X$ is called a periodic orbit if


*

*$x$ is a periodic function, and

*there exists a family $(f^t)_{t > 0}$ of functions $f^t \colon X \to X$ such that $f^{s+t}=f^t \circ f^s$ for all $s,t>0$, and $x(s+t)=f^t(x(s))$ for all $s \in \mathbb{R}$ and $t>0$.


So, for example, taking $X=\{0,1\}$, it is easy to see that the periodic function $\mathbf{1}_\mathbb{Q}$ is not a periodic orbit: e.g., if the family $(f^t)$ as above exists, then
$$ 0 = f^2(0) = f^\sqrt{2}(f^{2-\sqrt{2}}(0)) = f^\sqrt{2}(1) = f^\sqrt{2}(f^\sqrt{2}(0)) = f^{2\sqrt{2}}(0) = 1. $$
So my first question is:


Q1. Given a set $X$ (with cardinality at most that of the continuum) and a periodic orbit $x \colon \mathbb{R} \to X$, if $x$ is non-constant, does $x$ necessarily have a least period?


This question can alternatively be phrased more "directly" as follows: Given a family $(f^t)_{t > 0}$ of functions $f^t\colon X \to X$ satisfying $f^{s+t}=f^t \circ f^s$ for all $s,t > 0$, and a point $p \in X$, if there is a strictly decreasing sequence $t_n \to 0$ such that $f^{t_n}(p)=p$ for all $n$, does it necessarily follow that $f^t(p)=p$ for all $t>0$?


Q2. If the answer to Q1 is (in general) no, does it become yes if we additionally require $X$ to be a finite set?


 A: Two questions were asked
Concerning question I:
Here's an application for my favorite bijection!
Consider R the group of reals with addition and its subgroup Q of rationals.
The group R mod Q has the same cardinality as R does (I guess you need the axiom of choice here).
Let $\phi$ be a bijection from R mod Q to R.
Let $f^t$ be the conjugate by $\phi$ of the translation by t in R mod Q.
For all $t \in Q$ and all $x \in R$, $f^t(x)=x$.
Concerning question II: (Corrected from my previous wrong claim) For X finite, then your assumptions imply that all $f^t$ are the same map and are a projection (a solution $p$ of $p\circ p=p$). In particular, your claim is true if $X$ is finite.
Proof: let $n=|X|$, and notice that $f^t$ is for all k>1 the k-th iterate of a map $g:X\to X$ with $g=f^{t/k}$. Apply this to $k=n!$. I claim that $g^{n!}$ is a projection. Indeed consider any $x\in X$. Its orbit by $g$ consists in a tail of length $a\in\{0,\ldots,n-1\}$ followed by a cycle of length $b\in\{1,\ldots,n\}$, and $a+b\leq n$. So $g^{a}(x)$ has period $b$ dividing $n!$. As $n!\geq a$, the point $y=g^{n!}(x)$ is also a $g$-periodic point of period dividing $n!$, so it is fixed by $g^b$. Since $b$ divides $n!$, the point $y$ is fixed by $g^{n!}$. This proves the claim. Hence for all $t$, $f^t$ is a projection.
Now consider two positive reals $s<t$. Then $f^t=f^s\circ f^{t-s}$ so the image of $f^t$ is contained in the image of $f^s$. But the image of $f^t$ is the same as the image of $f^{t/k}$ for all $k$ because the latter is also a projection. Since there is some $k$ so that $t/k<s$, we get that the image of $f^s$ is contained in the image of $f^{t/k}$ i.e. in the image of $f^t$. Since projections are characterized by their images, we are done.
