I'd like to use the following fact as part of a proof on a problem set, but I'm not sure if it's true.
Let $P = \{x_1, x_2, ..., x_n\}$ be a partition of $[a, b]$, and suppose $s$ is constant on each open subinterval $(x_{k+1}, x_k)$. Does there exist a partition $P' = \{x', x' + k, x' + 2k, ..., x' + nk\}$ such that $s$ is constant on each open subinterval $(x' + ik, x' + (i + 1)k)$? i.e.: for every step function, does there exist a partition with "equally-spaced" elements?
If every element of $P$ is rational, I think the answer is yes -- rewrite all of the $x_i$s as $\frac{p_i}{q_i}$ and use something like $\frac{1}{\prod_{i = 1}^n q_i}$. If every element of $P$ is algebraic, I suspect it's possible to transform $P$ in a way that makes every element rational and use the same strategy, though I'm not sure exactly how to go about it. But for something like $P = \{e, \pi, 10\}$, it's not clear to me that there is any way to find a "chunk size" that divides both $\pi - e$ and $10 - \pi$.
Without loss of generality, I'll just consider partitions of $[0,1]$. Like you said, if the partition $0=x_0<\dots<x_n=1$ consists of only rational numbers, then it is possible to find an equidistributed subdivision. However, this is more than just sufficient: it is also necessary.
Suppose $P = \{x_0,\dots,x_n\}$ is some partitioning of $[0,1]$ and we can subdivide it into some $Q = \{0, k, 2k, \dots,1\}$ into equally spaced partitions. Let $N+1$ denote the length of $Q$, then $Nk=1$ shows that $k$ is rational. Since every element of $P$ lies in $Q$, this implies that every element of $P$ must be rational also.
Reparametrising to address partitions of $[a,b]$, we get that: