New Order #2: Turn My Way
JavaScript (ES6), 65 bytes
1-indexed.

n=&gt;{for(o=p=[k=1];o[k]|~-(i=p^k)&amp;i?k++:k=o[p=k]=!!n--;);return p}

Try it online!
Commented
n =&gt; {                  // n = index of requested term
  for(                  // for loop:
    o =                 //   o = storage object for the terms of the sequence
    p =                 //   p = last term found in the sequence
      [k = 1];          //   k = current term
    o[k] |              //   if k was already encountered
    ~-(i = p ^ k) &amp; i ? //   or (p XOR k) has more than 1 bit set:
      k++               //     increment k
    :                   //   else:
      k = o[p = k]      //     set o[k], set p to k
        = !!n--;        //     stop if n is equal to 0 or set k to 1; decrement n
  );                    // end of for()
  return p              // return p
}                       // end