Let $A=\{p_1,\ldots,p_m\}$, $B=\{q_1,\ldots,q_m\}$, and $A,B \subset \mathbb{R}^n$.
Is it true that there is an isometry that sends $A$ to $B$ iff Gramian matrices corresponding to $A$ and $B$ are similar and the similarity is achieved by permutation matrices?
Additionally, is there a similar condition that depends only on checking a (easily computable) property of a single matrix rather than trying to find if the two matrices are similar via permutation matrices?
I think this is a counterexample to the forward direction. Let $n = 2$, $A = \{ (1,0), (0,1) \}$, and $B = \{ (1,0), (1+ \sqrt{2}, 0) \}$. Rotating 135 degrees (clockwise) about the point $(1,0)$ is an isometry of $\mathbb{R}^2$ that takes the set $A$ to $B$. The Gram matrices of $A$ and $B$ are $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad \mathrm{and} \quad \begin{pmatrix} 1 & 1 + \sqrt{2} \\ 1 + \sqrt{2} & (1 + \sqrt{2})^2 \end{pmatrix}$$ The first matrix is invertible, but the second is not since it has determinant 0, so the two matrices cannot be similar.
You can generalize this example by finding isometries that take a set of linearly independent vectors to a set of linearly dependent vectors, because the Gram matrix of a collection of vectors is invertible if and only if the set of vectors is linearly independent.