= Solution
Set $f=p(x)e^{-x^2/2}$. Differentiating gives
$$
f''-x^2f=[p''-2xp'-p]e^{-x^2/2}.
$$
Thus the <polynomial> must satisfy $p''-2xp'-(\mu+1)p=0$. If it has degree $n$ and nonzero leading <coefficient>, comparison of highest powers gives $\mu=-(2n+1)$. For this value the equation becomes the <Hermite differential equation>
$$
p''-2xp'+2np=0.
$$
Write $p=\sum_{j=0}^na_jx^j$, with <coefficients> beyond $n$ zero. Comparing powers yields
$$
(j+2)(j+1)a_{j+2}+2(n-j)a_j=0.
$$
The equation at $j=n-1$ forces $a_{n-1}=0$, and downward recurrence removes all terms of the opposite parity and determines all same-parity terms from $a_n$. Taking $a_n=1$ produces the explicit <polynomial>
$$
p_n(x)=\sum_{r=0}^{\lfloor n/2\rfloor}
\frac{(-1)^r n!}{4^r r!(n-2r)!}x^{n-2r}.
$$
It satisfies every <coefficient> equation, so existence as well as uniqueness up to a nonzero scalar is proved. Hence
$$
\boxed{\mu_n=-(2n+1),\qquad f_n=p_ne^{-x^2/2}.}
$$
The monic $p_n$ is $2^{-n}$ times the usual physicists' <Hermite polynomial>.
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