I posed this question in a Facebook group for math problems:
What is the size of the largest set $S \subset \mathbb{R}^n$ for which there exists some norm $||\cdot||: \mathbb{R}^n \to \mathbb{R}_{\geq 0}$ on $\mathbb{R}^n$ (as a vector space over $\mathbb{R}$) such that $||p - q||$ has the same value for all distinct $p, q \in S$?
Under the Euclidean norm, the largest such sets are the $n+1$ vertices of a regular $n$-dimensional simplex. Other norms allow larger sets; the best result I know is the $2^n$ vertices of the unit hypercube under the uniform norm $||(x_1, \ldots, x_n)||_\infty = \max_{1 \leq i \leq n} |x_i|.$
Benjamin Gunby, a math Ph.D. student at Harvard, has proved an upper bound of $3^n$ with the following argument (paraphrased):
Let $||\cdot||$ be a norm. For $p \in \mathbb{R}^n$ and $r > 0$, let $B(p, r) = \{ q \in \mathbb{R}^n: ||q - p|| < r\}$ be the open ball with center $p$ and radius $r$. Let $V$ be the volume of a ball of radius $1$; the volume of a ball of radius $r$ is thus $r^n V$. Let $p_1, \ldots, p_k$ be $k$ points in $\mathbb{R}^n$ such that the distance between any two distinct points is $2$. By the triangle inequality, the balls $B(p_i, 1)$ are pairwise disjoint and contained in $B(p_1, 3)$. Thus, $\operatorname{Vol} \left(\bigcup_{i=1}^k B(p_i, 1)\right) = kV \leq \operatorname{Vol} B(p_1, 3) = 3^n V$, so $k \leq 3^n$.
Can either bound be improved upon? As a final note, $\mathbb{Q}^n$ admits arbitrarily large finite sets with all points equidistant, such as $\{1, \ldots, p-1\}^n$ under the $p$-adic uniform norm $||(q_1, \ldots, q_n)|| = \max_{1 \leq i \leq n} ||q_i||_p$, where $p$ is an arbitrary prime. (Volume is not a meaningful concept in $\mathbb{Q}^n$ under $p$-adic norms, so the argument for the upper bound does not apply.)