Here'sTwo 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 projectionsa projection (solutionsa 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.