Take $H=L^2(\mathbb{R}^n, e^{-|x|^2}dx)$ as your Hilbert space, the $L^2$ space with weight $e^{-|x|^2}$, so that polynomials are in the space (Hermite Polynomials are an ONB for this space).

Your map $(B, b) \mapsto \widehat{(B, b)}$ defines a group homomorphism from the group $T(n)$ of transformations $x \mapsto Bx+b$ into the group $\mathrm{GL}(H)$ of bounded linear isomorphisms of $H$. Now the differential of your map at the identity assigns to an element $x \mapsto Mx + b$ in $\mathfrak{t}(n)$ (i.e. $M \in \mathrm{End}(\mathbb{R}^n)$, $b \in \mathbb{R}^n$) the unbounded operator (infinitesimal generator)
$$ \widehat{(M, b)}f(x) = Df(x) \cdot(Mx + b).$$
I didn't think too long about this, but the generalized eigenfunctions of this operator should simply be certain polynomials. For example, if $c$ is a left eigenvector of $B$ with eigenvalue $\lambda \neq 0$ and we set $d = \langle c,b\rangle/\lambda$, then $f(x) = \langle c, x \rangle + d$ is and eigenvector to the eigenvalue $\lambda$. I think that if $M$ is diagonalizable, then $\widehat{(M, b)}$ has a discrete spectrum $\lambda_1, \lambda_2, \dots$ and a complete eigendecomposition, with each eigenfunction being a polynomial.

If now $B = e^M$, then $\widehat{(B, b)}$ has the same eigenfunctions with eigenvalus $e^{\lambda_1}, e^{\lambda_2}, \dots$.

*By the way: A functional is usually a function from a space to the scalars, so you do not define a functional but an operator. This confused me for a bit.*

\Edit: Assume $B = e^M$. Then $(M, b)$ is an element of the Lie algebra of $T(n)$. Let $U_\varepsilon = \exp(\varepsilon M, \varepsilon b)$ (Note this is the exponential in the Lie group $T(n)$, which is given on the matrix part by the usual matrix exponential, but on the translation part, $\exp(0, b)(x) = x + b$. $U_\varepsilon$ is given by the Baker-Campbell-Hausdorff formula in general, but its Taylor expansion is
$$\exp(\varepsilon M, \varepsilon b)(x) = x + \varepsilon M + \varepsilon b + O(\varepsilon^2).$$
Hence if $f$ is somewhat regular (say a polynomial at first),
$$\hat{U}_\varepsilon f(x) = f(x + \varepsilon M + \varepsilon b) + O(\varepsilon^2) = f(x) + \varepsilon Df(x)(M + b) + O(\varepsilon^2).$$
Now forming
$$f = \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} (U_\varepsilon f - f)$$
gives what I said. Notice that the limit is actually in $L^2$ with the given weight (if $f$ is a polynomial, the space of which is dense).