Let $X$ be a set whose cardinality is at most that of the continuum. Let $(f^t)_{t > 0}$ be a $(0,\infty)$-indexed family of functions $f^t\colon X \to X$ with the property that $f^{s+t}=f^t \circ f^s$ for all $s,t > 0$.

Suppose we have $x \in X$ and a strictly decreasing sequence $t_n \to 0$ such that $f^{t_n}(x)=x$ for all $n$. Does it necessarily follow that $f^t(x)=x$ for all $t>0$? If it does not follow in general, does it necessarily follow in the case that $X$ is a finite set?

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?