Let $M$ be a compact, connected $m$-dimensional Riemannian manifold, and let $n\in\mathbb{N}$. Can we always find distinct points $p_1,\dotsc,p_n\in M$ such that for $i=1,\dotsc,n$ the regions $A_i=\{p\in M;d(p,p_i)<d(p,p_j)\;\forall j\neq i\}$ all have the same volume?

I'm sure someone has studied this problem before but I haven't found any information about it. I suspect the statement is true, but I have no basis for it.

There are some cases where it is true:

• The case mentioned in the comments: there is an action by isometries on $M$ with some orbit having $n$ points.

• When $n=2$ the statement is a consequence of the intermediate value theorem: consider two disjoint arcs joining $p_1$ and $p_2$ and swap them using the arcs. In general, as $\mu(A_1),\dotsc,\mu(A_n)$ depend continuously on $p_1,\dotsc,p_n$ (see the remark below) maybe some argument using homology (degree theory/Brouwer's fixed point theorem/something like that) could work, but I have not found anything.

In case we can always find the $n$ points, here is a stronger version of the question:

Given some nonempty open set $U\subseteq M$, can we choose points $p_1,\dotsc,p_n\in U$ that divide $M$ into regions of equal volume?

Remark: If $M$ is a closed Riemannian manifold and $\mu$ is the measure associated to Riemannian volume, then the function $$M^n\setminus\{(p_1,\dotsc,p_n);p_i=p_j\text{ for some }i\neq j\}\to\mathbb{R}^n;(p_1,\dotsc,p_n)\mapsto(\mu(A_1),\dotsc,\mu(A_n))$$ is continuous.

Proof: Let $p_1,\dotsc,p_n$ be different and let $\varepsilon>0$. We want to find $\delta$ such that $\forall q_1,\dotsc,q_n$ with $d(q_i,p_i)<\delta$ we have $\lvert\mu(A_i)-\mu(B_i)\rvert<\varepsilon\;\forall i$ (where $A_i$, $B_i$ are the regions associated to the $p_i$ and the $q_i$ respectively). By my answer to Bisector of two points in a Riemannian manifold has measure $0$, the set of points $x\in M$ such that $d(x,p_i)=d(x,p_j)$ for some $i\neq j$ has measure $0$, so for some $\delta$ the set $X=\{x\in M;\lvert d(x,p_i)-d(x,p_j)\rvert<\delta\text{ for some }i\neq j\}$ has measure $<\varepsilon$. So if $d(q_i,p_i)<\frac{\delta}{2}$ for all $i$, then $A_i\setminus X\subseteq B_i\subseteq A_i\cup X$, thus $\lvert\mu(A_i)-\mu(B_i)\rvert<\varepsilon$ for all $i$, as we wanted.