Consider the $n=1$ case, the generalization to $n\in \mathbb N$ should be easy. Also, note that the following is not completely rigorous, because we do not talk about domains and so on. However, I find the proof quite easy to follow. It can be generalized to normal $B$ in a straightforward manner.

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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

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

This implies that for $b_\ell\neq 0$, $a^\dagger |b_\ell\rangle$ is an eigenstates of $B$ with eigenvalue $1/b_\ell$. Now observe that $b_\ell=0$ would imply that $a^\dagger |b_\ell\rangle=0$, i.e. 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, so $b_\ell\neq 0$ for all $b_\ell$. 

Similarly to $(1)$, we obtain that $a|b_\ell\rangle$, $a^\dagger a|b_\ell\rangle$ and $a a^\dagger |b_\ell\rangle$ are eigenvectors of $B$ with eigenvalue $1/b_\ell$. With the help of the CCR, this readily implies that 
$$ B|b_\ell\rangle=b_\ell |b_\ell\rangle = B I |b_\ell\rangle= B(aa^\dagger -a^\dagger a)|b_\ell\rangle = 1/b_\ell |b\rangle \tag 2 \quad ,$$

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. $\tag*{$\square$}$