Consider the $n=1$ case, the generalization to $n\in \mathbb N$ should be easy. It can be generalized to normal $B$ in a straightforward manner.

To start, since $B$ is hermitian, it admits a complete orthonormal basis of eigenvectors. Let $B|b_\ell\rangle=b_\ell|b_\ell\rangle$ such that $B=\sum\limits_{\ell} b_\ell|b_\ell\rangle\langle b_\ell|$. Then by assumption we have

$$Ba^\dagger B|b_\ell\rangle=b_\ell Ba^\dagger |b_\ell\rangle=a^\dagger |b_\ell\rangle \quad   \tag 1 \quad .$$

 Now observe that $a^\dagger |b_\ell\rangle=0$ implies that $|b_\ell\rangle$ is an eigenvector of $a^\dagger$. It is [well-known](https://physics.stackexchange.com/questions/155852/is-there-a-simple-way-of-finding-the-eigenstates-of-the-creation-and-annihilatio) however that this operator does not have any eigenvectors. This means that $b_\ell\neq 0$ and also that $a^\dagger |b_\ell\rangle$ is an eigenstates of $B$ with eigenvalue $1/b_\ell$.

Next, we consider the vectors $a|b_\ell\rangle$, $a^\dagger a|b_\ell\rangle$ and $a a^\dagger |b_\ell\rangle$. For the last one, we find that $aa^\dagger|b_\ell\rangle=(I+a^\dagger a)|b_\ell\rangle =0$ implies $a^\dagger a|b_\ell\rangle= - |b_\ell\rangle$, but since $a^\dagger a\geq 0$, this is not possible. Using $(1)$, it then follows that

$$aa^\dagger |b_\ell\rangle= BaBa^\dagger |b_\ell\rangle= 1/b_\ell Baa^\dagger |b_\ell\rangle \tag 2\quad , $$

which shows that $aa^\dagger |b_\ell\rangle$ is an eigenvector of $B$ with eigenvalue $1/b_\ell$.

To proceed, notice that $a|b_\ell\rangle=0$ as well as $a^\dagger a|b_\ell\rangle=0$ implies that $|b_\ell\rangle=|0\rangle$ up to a phase (uniqueness of vacuum/ non-degeneracy of the number operator). Then as long as $|b_\ell\rangle \neq |0\rangle$, a computation similar to $(1)$ and $(2)$ shows that both vectors $a|b_\ell\rangle$ and $a^\dagger a|b_\ell\rangle$ are eigenvectors of $B$ with eigenvalue $1/b_\ell$.

With the help of the CCR, this readily implies that 
$$ B|b_\ell\rangle=b_\ell |b_\ell\rangle = B I |b_\ell\rangle= B(aa^\dagger -a^\dagger a)|b_\ell\rangle = 1/b_\ell |b_\ell\rangle \tag 2 \quad ,$$

for all $|b_\ell\rangle$ (including possibly $|0\rangle$), meaning that $b_\ell=1/b_\ell$ or $b_\ell=\pm 1$ and hence $B^2=I$, i.e. $B=B^\dagger=B^{-1}$. 

This in turn implies that $B$ commutes with both $a$ and $a^\dagger$ and by Schur's lemma it follows that $B=\pm I$, concluding the proof.