Only $B_2$ splits into an odd number of equivalence classes, namely, three: $\{(0,0)\}$, $\{(0,1), (1,0)\}$ and $\{(1,1)\}$. If $n > 2$, then $B_n$ always splits into an even number of classes, as we are presently going to show.

Call two $n$-tuples of $B_n$ conjugate if one is obtained from the other by replacing its zeroes by ones and its ones by zeroes. With each class in $B_n$ we may associate the class of conjugate $n$-tuples. Thus, if there are no self-conjugate classes, the number of classes is even. This is clearly the case if $n$ is odd, for the $n$-tuples in a self-conjugate class must have the same number of zeroes and ones, so the total number of entries is even.

Write $n = 2^m n'$, where $m$ and $n'$ are both non-negative integers, and $n'$ is odd, and induct on $m$. The base case, $m=0$, is clear by the preceding, so let $m \ge 1$.

Let $n_k$ be the number of $k$-element classes in $B_n$. Clearly, if $n_k \neq 0$, then $k$ divides $n$, so $2^n = |B_n| = \sum_{k|n} k n_k = \sum_{k|n/2} k n_k + \sum_{2^m|k} k n_k = \sum_{k|n/2} k n_k + \sum_{d|n'} 2^m d n_{2^m d}$.

If $k$ divides $n/2$, then an $n$-tuple belonging to a $k$-element class in $B_n$ can be obtained only by iterating an $n/2$-tuple in $B_{n/2}$, so the first sum in the last expression above is $2^{n/2}$.

Consequently, $\sum_{d|n'} d n_{2^m d} = 2^{n/2-m}(2^{n/2}-1)$. Notice that $n/2-m = 2^{m-1}n' - m$ is positive, unless $m=1$ or $m=2$, and $n'=1$, to infer that, with these exceptions, $\sum_{d|n'} d n_{2^m d}$ is even. Since $n'$ is odd, so are all its divisors, and it follows that $\sum_{d|n'} n_{2^m d}$ is even, showing that the total number of $k$-element classes in $B_n$ is even, whenever $k$ is divisible by $2^m$.

If $k$ is not divisible by $2^m$, then the total number of $k$-element classes in $B_n$ is the same as the total number of classes in $B_{n/2}$ which is even by the induction hypothesis, save when $n=4$. This corresponds to an exceptional case above; the other exceptional case is $n=2$, when the number of classes is clearly three.

Finally, the case $n=4$ is dealt with by explicit construction of the (six) classes $B_4$ splits into: There are two singleton classes, represented by $(0,0,0,0)$ and $(1,1,1,1)$, respectively; there is only one doubleton class, represented by $(0,1,0,1)$; and there are three 4-element classes, represented by $(0,0,0,1)$, $(0,0,1,1)$ and $(0,1,1,1)$, respectively.