# [Python], <sup><s>262</s> \$\cdots\$ <s>307</s></sup> 305 bytes

Saved a whopping 19 bytes thanks to [dingledooper](https://codegolf.stackexchange.com/users/88546/dingledooper)!!! 

Added 118 bytes to fix a bug kindly pointed out by [xnor](https://codegolf.stackexchange.com/users/20260/xnor), [Peter Kagey](https://codegolf.stackexchange.com/users/53884/peter-kagey) and [l4m2](https://codegolf.stackexchange.com/users/76323/l4m2).

<!-- language-all: lang-python -->

 lambda l,R=range:(n:=len(l))<2or(d:=len(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(len(l[0])))and len({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

[Try it online!][TIO-ka53jn0q]

[Python]: https://docs.python.org/3.8/
[TIO-ka53jn0q]: https://tio.run/##nVLBbqMwEL3nK0a9ZIZ1VkB21QbVe@ylt2hvLqpIIK0jMMg4UtMo3561DUnIpj20GIk3M@@N3xg3W/Naq@ldow8r/nQos2qRZ1CyOdeZeikSVAkvC4Ul0X1ca8y7cCEVKqLJlDKVY5PwttamyFG0mwrxbbKlIIhXtYY3tgWp4F02KNmayOWky5Tg4NrDlIhzIYNGqPREmGP@I/b85y6MgyAPqsy8/lzW1QJzJolSt/1jwufoTYrQtnKWwIU7b8ZbKoVMxXrYXHVevIFHV5uooGcdvXc1uhDt7TGMZNXYccF5OZiiNS1wECOwjxAYMZgxuLUvpQz@6k2RsnONGGDMrishA/v@dmX7jXoYnbNRn/2gp@OEZ@URuuisjJzyISvbr0jDnnAtDS@kA/6xS/R9adgbHnQ5wcFE4X/DXR7LzBdmPSd28ItH9@l5XamuBvog8ekvCLtFzHKjAe6jE7ZrqE9HmuNm/LSJb38txww6GE3HNHIX1rDCXVl/OxO/WctXaMjDRktlcHWzM3uY/IFdu4edFi3nRbq/ocM/ "Python 3.8 (pre-release) – Try It Online" 


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.