Python, 262 \$\cdots\$ 307305 305303 bytes
Saved a whopping 19 bytes thanks to dingledooper!!!
Added 118 bytes to fix a bug kindly pointed out by xnor, Peter Kagey and l4m2.
lambda l,R=range,L=len:(n:=len=L(l))<2or(d:=len=L(bin(n))-3)and(p:=sorted([sum((x-y)**2for x,y in zip(i,j))for i in l for j in l]))==[i*p[n]for i in R(d+2)for _ in R(2**d*math.comb(d,i))]and(K:=R(lenL(l[0])))and lenL({sum(([sum(l[i][j]for i in R(n))for j in K][j]-n*l[i][j])**2for j in K)for i in R(n)})<2 import math Inputs a list of points and returns True/False.
How
Calculates the square of the distances between all possible pairs of points (including self-pairs and both \$(p_i,p_j)\$ and \$(p_j,p_i)\$ for all points \$p_j\$ and \$p_i\$ where \$i\neq j\$) and normalises them by the smallest non-zero square distance. For an \$n\$-cube we should then see a pattern of integers \$i = 0,1,\dots, n\$ each occurring \$2^{n}{n\choose i}\$ times. This corresponds with the \$0\$s for all the self-pairs, and the square of the lengths of all the sides being \$a^2\$, and the square of the lengths of all the diagonals being \$2a^2, 3a^2,\dots, na^2\$.
Correction
Also checks that the given vertices are all equidistant from the centre of mass.
