Test if a Gray code is Beckett
Husk, 9 bytes
Λ≡½hÖ▼Ẋz-

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Explanation
If we compute differences from each binary sequence to the next with an element-by-element subtraction, we will get a series of results each containing a single -1/+1 on the flipped bit and a 0 on all other bits. In order for this Gray code to be Beckett, the subsequence of -1 lists must be identical to the subsequence of +1 lists (except for the sign).
As noted by Unrelated String we don't need to check the last element against the first, since it will always consist of a single 1 turning into a 0.
 Λ≡½hÖ▼Ẋz-    Input: list of binary sequences
       Ẋ      For each pair of adjacent sequences:
        z-       subtract the first from the second element by element
     Ö▼       Sort by the minimum (in place): all the -1 sequences move to the beginning
    h         Discard the last (would match the final sequence turning into the first)
   ½          Group the sequences into two lists of equal length 
                (the first will contain the -1s, the second the +1s)
 Λ            Check that the two elements
  ≡           Have the same distribution of zeros/nonzeros