Yes, once you have the equation
$$
\sum_{n}\mathscr{F}(e^{-x^2/2}H_n(x))\frac{t^n}{n!}
= \sum_{n}\sqrt{2\pi}(-i)^n H_n(\xi)e^{-\xi^2/2}\frac{t^n}{n!},
$$
then you can fix $\xi$, and view this as a power series equation that holds for all $t$. Therefore, for this fixed $\xi$, the power series coefficients must be identical, which leads to
$$
\mathscr{F}(e^{-x^2}H_n(x))=\sqrt{2\pi}(-i)^n H_n(\xi)
$$
The left side implicitly depends on $\xi$, because it is the transform variable. The transform is continuous in $\xi$ because of the exponentially decaying nature of the function being transformed. So
$$
\mathscr{F}(e^{-x^2}H_n(x))(\xi)=\sqrt{2\pi}(-i)^n H_n(\xi),\;\;\xi\in\mathbb{R}.
$$
Omitting the arguments of the functions, these functions are equal:
$$ \mathscr{F}(e^{-x^2}H_n(x)) = \sqrt{2\pi}(-i)^n H_n
$$
We can prove that:
$$I(k,a)=\int_{-\infty}^{\infty}H_n(ax)e^{-\frac{x^2}{2}}e^{-ikx}dx=\sqrt{2\pi}(-i)^n\Big(2a^2-1\Big)^{n/2}H_n{\Big(\frac{ak}{\sqrt{2a^2-1}}\Big)}e^{-\frac{k^2}{2}}$$
This shows that when $a=1$ we recover the desired result, that the Hermite polynomials are eigenfunctions of the Fourier transform. This also allows to compute the second Fourier transform in question since:
$$\int_{-\infty}^{\infty}H_n(k)e^{-\frac{k^2}{2b^2}}e^{ikx}\frac{dk}{2\pi}=\int_{-\infty}^{\infty}H_n(x)e^{-\frac{x^2}{2b^2}}e^{ikx}\frac{dx}{2\pi}=\frac{b}{2\pi}\int_{-\infty}^{\infty}H_n(b x)e^{-\frac{x^2}{2}}e^{ikb x}dx=(-1)^n\frac{b}{2\pi}I(kb,b)$$
and therefore we obtain the slightly more general result, valid for complex $b$ in general for which the integral converges:
$$I(x,a,b)=\int_{-\infty}^{\infty}H_n(ak)e^{-\frac{k^2}{2b^2}}e^{ikx}dk=i^nb\sqrt{2\pi}(2a^2b^2-1)^{n/2}H_n\Big(\frac{a~b^2 ~x}{\sqrt{2a^2b^2-1}}\Big)e^{-b^2x^2/2}$$
from which we finally obtain that:
$$\mathcal{F^{-1}}(\sqrt{2\pi}(-i)^nH_n(k)e^{-(1+iz)k^2/2})=\frac{(-1)^n}{\sqrt{1+iz}}\Big(\frac{1-iz}{1+iz}\Big)^{n/2}H_n\Big(\frac{x}{\sqrt{1+z^2}}\Big)e^{-\frac{x^2}{2(1+iz)}}$$
Note for calculation of $I$ to be added soon.
$\textbf{EDIT:} ~~\small\text{Calculation of $I(k,a,b)$ valid for all $b\in\mathbb{C}$}$
First, write $H_n(x)=\sum_{l}c_{nl}x^l$. Substitute this in and perform the integrals:
$$I(k,a,b)=\sum_{l}c_{nl}a^l\int_{-\infty}^{\infty}x^le^{-x^2/2b^2}e^{-ikx}dx\\=b\sqrt{2\pi}\sum_{l}c_{nl}a^l\Big(i\frac{d}{dk}\Big)^le^{-k^2b^2/2}\\=b\sqrt{2\pi}\sum_{l}c_{nl}(iab)^le^{-k^2b^2/2}\Big[e^{k^2b^2/2}\Big(\frac{d}{d(kb)}\Big)^le^{-k^2b^2/2}\Big]\\=b\sqrt{2\pi}\sum_{l}c_{nl}(-iab)^le^{-k^2b^2/2}He_n(kb)$$
where $He_n(x)$ are the probabilists Hermite's polynomials as defined on the wikipedia page.
Now utilize the represenations in terms of differential operators$$He_n(ax)=a^ne^{-D^2/2a^2}x^n, H_n(ax)=(2a)^ne^{-D^2/4a^2}x^n, D\equiv\frac{d}{dx}$$ repeatedly to rewrite again as:
$$\begin{align}I(k,a,b)&=b\sqrt{2\pi}e^{-k^2b^2/2}e^{-D^2/2b^2}\sum_{l}c_{nl}(-iab^2k)^l\\&=b\sqrt{2\pi}e^{-k^2b^2/2}e^{-D^2/2b^2}H_n(-iab^2k)\\&=b\sqrt{2\pi}e^{-k^2b^2/2}e^{-D^2/2b^2}(2iab^2)^n e^{D^2/4a^2b^4}k^n\\&=(2iab^2)^n e^{-k^2b^2/2}b\sqrt{2\pi}e^{-k^2b^2/2}e^{-D^2(1/2b^2-1/4a^2b^4)}k^n\\&=b\sqrt{2\pi}e^{-k^2b^2/2}\Big(-iab\sqrt{2-\frac{1}{a^2b^2}}\Big)^nH_n\Big(\frac{kb}{\sqrt{2-\frac{1}{a^2b^2}}}\Big)\end{align}$$
which upon some basic algebra yields the quoted result:
$$I(k,a,b)=b\sqrt{2\pi}(-i)^n\Big(2a^2b^2-1\Big)^{n/2}H_n\Big(\frac{kab^2}{\sqrt{2a^2b^2-1}}\Big)e^{-k^2b^2/2}$$
Best Answer
An alternative demonstration may use a proof by induction based on the recurrence relation $H_n(x) = xH_{n-1}(x) - H_{n-1}'(x)$, which itself implies $\varphi_n(x) = \frac{x\varphi_{n-1}(x) - \varphi_{n-1}'(x)}{\sqrt{2n}}$. However, we must be careful with the notation, because it can lead to an actual muddle in the present case.
For recall, we would like to prove that $\mathfrak{F}[\varphi_n(x)](k) = (\color{red}{-i})^n\varphi_n(k)$. The base case $n = 0$ is easily checked, since $\varphi_0$ is a gaussian, which is mapped to another gaussian in the Fourier space. Then we assume the wanted relation for $n-1$, so that the induction step gives : $$ \begin{array}{rcl} \mathfrak{F}[\varphi_n(x)](k) &=& \displaystyle \mathfrak{F}\left[\frac{x\varphi_{n-1}(x) - \varphi_{n-1}'(x)}{\sqrt{2n}}\right](k) \\ &=& \displaystyle \frac{1}{\sqrt{2n}}\left(i\frac{\mathrm{d}}{\mathrm{d}k}\mathfrak{F}[\varphi_{n-1}(x)](k) - ik\mathfrak{F}[\varphi_{n-1}(x)](k)\right) \\ &=& \displaystyle \frac{-i}{\sqrt{2n}}\left(k(-i)^{n-1}\varphi_{n-1}(k) - \frac{\mathrm{d}}{\mathrm{d}k}(-i)^{n-1}\varphi_{n-1}(k)\right) \\ &=& \displaystyle (-i)^n\frac{k\varphi_{n-1}(k) - \varphi_{n-1}'(k)}{\sqrt{2n}} \\ &=& \displaystyle (-i)^n\varphi_n(k) \end{array} $$