This is not a complete answer (and I'm sure there has to be a counterexample in some well known space), but here is an idea of how to change distances between points of a manifold by varying the metric. It seems that you can use it to modify the metric of $\mathbb{S}^3\subseteq\mathbb{R}^4$ so that it has $6$ equidistant points.

The idea is the following:

Take $P_1,P_2,P_3,P_4,P_5$ points in $\mathbb{S}^3$ which are equidistant respect to the usual metric, let $Q$ be the point of $\mathbb{S}^3$ opposite to $P_5$. Let $\gamma_{i,j}$ be the geodesic arc from $P_i$ to $P_j$ with the usual metric, and $d_0=d_g(P_1,P_2)$, with $g$ being the usual metric in $\mathbb{S}^3$.

Now consider the set $\displaystyle K=\bigcup_{i,j=1,\dots,5}\gamma_{i,j}$, and a small nhood of $K$, $U$. Multiplying $g$ by an adequate scalar function you can obtain a metric $g'$ such that, for some $\varepsilon>0$:

• $g'=\varepsilon g$ in $K$.
• $g'\geq\varepsilon g$ in $\mathbb{S}^3$.
• $g'=g$ outside of $U$.

This way $d_{g'}(P_i,P_j)=\varepsilon d_0$ for $i,j=1,\dots,5$, and choosing $\varepsilon$ adequately, we can get $d_{g'}(Q,P_i)=d_{g'}(P_i,P_j)$ for $i=1,2,3,4$.

So we have made the distances between the $P_i$ small by reducing the metric around the geodesics between them.

To make the distance from $P_5$ to $Q$ equal to the others, you can take a $g$-geodesic arc from $P_5$ to $Q$ which is disjoint from the arcs between the $P_i$ and reduce the metric around that arc too.

Anyways it's not obvious that the distances between the points are going to end up being what you want them to be. It seems to me that using this kind of technique and a more elaborate argument you could, given any $n$-manifold $M$ with $n\geq3$, points $P_1,\dots,P_k\in M$ and positive numbers $m_{i,j}$ for $1\leq i<j\leq k$ satisfying the triangle inequalities, create a metric in $M$ such that $m_{i,j}=d(P_i,P_j)$.

$K(M,g)$ depends on the metric, as shown by this question, which implies that we can change the metric of $\mathbb{R}^3$ so it has as many points pairwise at distance $1$ as we want.

This is not a complete answer (and I'm sure there has to be a counterexample in some well known space), but here is an idea of how to change distances between points of a manifold by varying the metric. It seems that you can use it to modify the metric of $\mathbb{S}^3\subseteq\mathbb{R}^4$ so that it has $6$ equidistant points.

The idea is the following:

Take $P_1,P_2,P_3,P_4,P_5$ points in $\mathbb{S}^3$ which are equidistant respect to the usual metric, let $Q$ be the point of $\mathbb{S}^3$ opposite to $P_5$. Let $\gamma_{i,j}$ be the geodesic arc from $P_i$ to $P_j$ with the usual metric, and $d_0=d_g(P_1,P_2)$, with $g$ being the usual metric in $\mathbb{S}^3$.

Now consider the set $\displaystyle K=\bigcup_{i,j=1,\dots,5}\gamma_{i,j}$, and a small nhood of $K$, $U$. Multiplying $g$ by an adequate scalar function you can obtain a metric $g'$ such that, for some $\varepsilon>0$:

• $g'=\varepsilon g$ in $K$.
• $g'\geq\varepsilon g$ in $\mathbb{S}^3$.
• $g'=g$ outside of $U$.

This way $d_{g'}(P_i,P_j)=\varepsilon d_0$ for $i,j=1,\dots,5$, and choosing $\varepsilon$ adequately, we can get $d_{g'}(Q,P_i)=d_{g'}(P_i,P_j)$ for $i=1,2,3,4$.

So we have made the distances between the $P_i$ small by reducing the metric around the geodesics between them.

To make the distance from $P_5$ to $Q$ equal to the others, you can take a $g$-geodesic arc from $P_5$ to $Q$ which is disjoint from the arcs between the $P_i$ and reduce the metric around that arc too.

Anyways it's not obvious that the distances between the points are going to end up being what you want them to be. It seems to me that using this kind of technique and a more elaborate argument you could, given any $n$-manifold $M$ with $n\geq3$, points $P_1,\dots,P_k\in M$ and positive numbers $m_{i,j}$ for $1\leq i<j\leq k$, create a metric in $M$ such that $m_{i,j}=d(P_i,P_j)$.

This is not a complete answer (and I'm sure there has to be a counterexample in some well known space), but here is an idea of how to change distances between points of a manifold by varying the metric. It seems that you can use it to modify the metric of $\mathbb{S}^3\subseteq\mathbb{R}^4$ so that it has $6$ equidistant points.

The idea is the following:

Take $P_1,P_2,P_3,P_4,P_5$ points in $\mathbb{S}^3$ which are equidistant respect to the usual metric, let $Q$ be the point of $\mathbb{S}^3$ opposite to $P_5$. Let $\gamma_{i,j}$ be the geodesic arc from $P_i$ to $P_j$ with the usual metric, and $d_0=d_g(P_1,P_2)$, with $g$ being the usual metric in $\mathbb{S}^3$.

Now consider the set $\displaystyle K=\bigcup_{i,j=1,\dots,5}\gamma_{i,j}$, and a small nhood of $K$, $U$. Multiplying $g$ by an adequate scalar function you can obtain a metric $g'$ such that, for some $\varepsilon>0$:

• $g'=\varepsilon g$ in $K$.
• $g'\geq\varepsilon g$ in $\mathbb{S}^3$.
• $g'=g$ outside of $U$.

This way $d_{g'}(P_i,P_j)=\varepsilon d_0$ for $i,j=1,\dots,5$, and choosing $\varepsilon$ adequately, we can get $d_{g'}(Q,P_i)=d_{g'}(P_i,P_j)$ for $i=1,2,3,4$.

So we have made the distances between the $P_i$ small by reducing the metric around the geodesics between them.

To make the distance from $P_5$ to $Q$ equal to the others, you can take a $g$-geodesic arc from $P_5$ to $Q$ which is disjoint from the arcs between the $P_i$ and reduce the metric around that arc too.

Anyways it's not obvious that the distances between the points are going to end up being what you want them to be. It seems to me that using this kind of technique and a more elaborate argument you could, given any $n$-manifold $M$ with $n\geq3$, points $P_1,\dots,P_k\in M$ and positive numbers $m_{i,j}$ for $1\leq i<j\leq k$ satisfying the triangle inequalities, create a metric in $M$ such that $m_{i,j}=d(P_i,P_j)$.