Apologies if this post will lack rigour, I'm only a humble physicist. Given a density operator $\hat\rho$ (i.e. a linear, positive self-adjoint operator with unity trace), one can represent it as the integral $$\hat\rho = \frac{1}{(2\pi)^m}\int_{\mathbb C^m} d^m\xi \text{ tr}[\hat\rho \hat D_{\xi}]\hat D_{-\xi} $$ performed over all the complex components $\xi_1,...,\xi_m$ of the vector $\xi$. The operator $\hat D_\xi$ is known as the displacement operator, but we don't really need to know how it is defined here. What I care about is how it transform under a certain unitary operation: suppose for instance that $\Phi_G(\hat D_{\xi})=\hat D_{G\xi}$ for some matrix $G$. Then by linearity $$\Phi_G(\hat\rho)= \frac{1}{(2\pi)^m}\int_{\mathbb C^m} d^m\xi \text{ tr}[\hat\rho \hat D_{\xi}]\hat D_{-G \xi}.$$ Here's the crux of the question. According to this representation, the operator $\hat\rho$ is fully determined by the quantity $\text{tr}[\hat\rho \hat D_\xi]$, the so-called characteristic function. In order to find the characteristic function of $\Phi_G(\hat\rho)$, I need to perform a change of variables $\xi'=G\xi$. However, this amounts to a change of variables inside an $m$-dimensional complex space, and I'm absolutely not sure that the usual results hold. For instance, if things were real I would note that for each component, $\xi_i'=G_i^{\ k} \xi_k$ giving the Jacobian $\partial_j\xi_i'=G_{ij}$ and hence
$$\Phi_G(\hat\rho) \overset{?}{=}\frac{1}{(2\pi)^m}\int_{\mathbb C^m} d^m\xi' \left|\det G\right|\text{ tr}[\hat\rho \hat D_{G^{-1}\xi}]\hat D_{-\xi'}. $$ What changes, if anything, in the complex scenario?