= Solution
Consider $B_n$ on the <finite-dimensional vector space> $E=\mathbb R[x]_{\le d}$ and use the monomial basis. For $j\ge1$, a <binomial distribution> and the <falling factorial> moment formula give
$$
B_n(x^j)=\frac1{n^j}\sum_{\ell=0}^j S(j,\ell)(n)_\ell x^\ell,
$$
where $S(j,\ell)$ is the <Stirling number of the second kind> and $(n)_\ell=n(n-1)\cdots(n-\ell+1)$. This follows by expanding $K^j=\sum_\ell S(j,\ell)(K)_\ell$ and using $\mathbb E(K)_\ell=(n)_\ell x^\ell$. Also $B_n(1)=1$.
Each monomial maps to a polynomial of no larger degree. The resulting matrix is triangular with diagonal entries $1$ and $(n)_j/n^j$, $1\le j\le d$. When $n\ge d$ and $n\ge1$, all entries are positive. Therefore
$$
\boxed{B_n(E)\subseteq E,\qquad B_n:E\to E\text{ is invertible for }n\ge\max(1,d).}
$$
For constant polynomials the conclusion is immediate. Taking positive indices avoids the undefined sample expression $k/n$ at $n=0$.
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