Take a (holomorphic) Cartan connection $E \to M$ modelled on a (complex) homogeneous space $(X,G)$, say $X=G/H$. Let $\omega$ be the Cartan connection on $E$. Every (holomorphic) vector field $v$ on $M$ is represented by a unique (holomorphic) $H$-equivariant function $f \colon E \to \mathfrak{g}/\mathfrak{h}$. Compute $df + \omega f = f' \omega$ for a unique $f' \colon E \to \mathfrak{g}/\mathfrak{h} \otimes (\mathfrak{g}/\mathfrak{h})^*$. The differential equation we need to have $v$ be an infinitesimal symmetry is that $f'$ is valued in $\mathfrak{g}$, acting as a linear map on $\mathfrak{g}/\mathfrak{h}$. So $Q=E \times^H W$$Q^0=E \times^H W$ where $W=(\operatorname{End} \mathfrak{g}/\mathfrak{h})/\mathfrak{g}$.
But then we need higher order conditions. Define $\mathfrak{g}^{(-1)}=\mathfrak{g}/\mathfrak{h}$. Define $\mathfrak{g}^{(0)}=\mathfrak{g}$. Define the prolongations $\mathfrak{g}^{(k)}$ to be the constant coefficient tensors $\phi \in \operatorname{Sym}^{k+1} (\mathfrak{g}/\mathfrak{h})^* \otimes \mathfrak{g}/\mathfrak{h}$ for which $\phi(v,*,\dots,*) \in \mathfrak{g}^{(k-1)}$. Define $f^{(-1)}=f$. Define $f^{(k)} \in \operatorname{End}(\mathfrak{g}/\mathfrak{h})^{(k)}$ by $df^{(k)}+\omega f^{(k)} = f^{(k+1)} \omega$. Define $Q^{(k)}=\operatorname{End}(\mathfrak{g}/\mathfrak{h})^{(k)}/\mathfrak{g}^{(k)}$. Then I think we get the differential operator $f \mapsto f^{(k)} + \mathfrak{g}^{(k)} \in Q^{(k)}$. For a flat $(X,G)$-geometry, these operators all vanish just when a vector field lies in the Lie algebra $\mathfrak{g}$, i.e. a local symmetry vector field. But what a mess. My choice of $k$ is maybe not what it should be, because it is one less than the order of the operator, but that is probably not unusual in the literature.