For a function $f: \mathbb R \to \mathbb R$ which is square-integrable w.r.t the $e^{-x^2}dx$, its $n$th Hermite coefficient is defined by $$ c_n(f) := \int_{-\infty}^\infty e^{-x^2}H_n(x)f(x)dx, $$ where $H_n$ is the (physicist's) $n$th Hermite polynomial.
For any $\alpha > 0$, define a new function $f_\alpha:\mathbb R \to \mathbb R$ by $f_\alpha(x) := f(\alpha x)$.
Question. Is there any interesting relationship between the the Hermite coefficients of $f_\alpha$ and those of $f$ ?
Example: positive-homogeneous functions
Suppose $f$ is positive homogeneous of order $p \ge 0$, i.e $f_\alpha(x) = \alpha^pf(x)$ for all $\alpha > 0$. Then $c_n(f_\alpha) = \alpha^p c_n(f)$ for all $n$.
To find the connection, look at the "Hermite multiplication theorem" $$ H_n(\alpha x) = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \frac{1}{k!(n-2k)!}\alpha^{n-2k}\left(\alpha^2-1\right)^{k}H_{n-2k}(x). $$
We have $$ f(x) = \sum_{n=0}^\infty c_nH_n(x) $$ therefore $$ f(\alpha x) = \sum_{n=0}^\infty c_nH_n(\alpha x) \\ = \sum_{n=0}^\infty c_n \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \frac{1}{k!(n-2k)!}\alpha^{n-2k}\left(\alpha^2-1\right)^{k}H_{n-2k}(x). $$ But what we need is a form $$ f(\alpha x) = \sum_{m=0}^\infty d_m H_m(x). $$ We can split the summation over $n$ into odd and even sums, however we can keep the notation lighter by defining $p=\operatorname{mod}(n,2)$, the parity of the index $n$. Let $n=2m+p$
$$ f(\alpha x)= \sum_{m=0}^\infty c_{2m+p} \sum_{k=0}^{m} \frac{\alpha^{2m-2k+p}\left(\alpha^2-1\right)^{k}}{k!(2m-2k+p)!}H_{2m-2k+p}(x). $$
Let $l=m-k$ to get it into the required form $$ f(\alpha x)= \sum_{m=0}^\infty c_{2m+p} \sum_{l=0}^{m} \frac{\alpha^{2l+p}\left(\alpha^2-1\right)^{m-l}}{(m-l)!(2l+p)!}H_{2l+p}(x). $$
Now, if the sum is absolutely convergent we can switch the order of summation from $$m\in\left[0,\infty\right]\\l\in\left[0,m\right] $$ to $$ l\in\left[0,\infty\right]\\m\in\left[l,\infty\right] $$ leading to $$ f(\alpha x)= \sum_{l=0}^\infty \sum_{m=l}^{\infty}c_{2m+p} \frac{\alpha^{2l+p}\left(\alpha^2-1\right)^{m-l}}{(m-l)!(2l+p)!}H_{2l+p}(x). $$ in which case we can read off the coefficients $$ f(\alpha x) = \sum_{l=0}^\infty d_{2l+p} H_{2l+p}(x) $$ as $$ d_{2l+p} = \sum_{m=l}^\infty c_{2m+p}\frac{\alpha^{2l+p}\left(\alpha^2-1\right)^{m-l}}{(m-l)!(2l+p)!} $$ where $p=\operatorname{mod}(l,2)$ is the parity of $l$.