Is there a Riemannian metric $\tilde g$ on $\mathbb R^d$ such that $$\tag{1} \Delta_{\tilde g}=e^f(\Delta +1),$$ for some $f\in C^\infty(\mathbb R^d)$? Here $\Delta=\partial_{x_1}^2+\ldots+\partial_{x_d}^2.$ (Answer: no, as (1) fails on constants. See Edit below).

If there is such a $\tilde g$, it cannot be conformal to the standard Euclidean metric $g=\delta_{ij}$. Indeed, if $\tilde g = e^{2\phi}g$, then $$\Delta_{\tilde g} = e^{-2\phi} \left(\Delta + (d-2)g^{ij}\frac{\partial \phi}{\partial x_j}\frac{\partial}{\partial x_i}\right),$$ and either $d=2$, or the second summand in the round brackets is constant only in the trivial case $\nabla \phi=0$. In both cases (1) cannot be satisfied.

**EDIT**.
The equation (1) cannot hold verbatim, as it clearly fails on constant functions (thanks Terry Tao for this remark). Instead, let us consider
$$
\tag{1b} L_{\tilde g} = e^f(\Delta +1), $$
where
$$
L_{\tilde g}=\frac{d-1}{4(d-2)} \Delta_{\tilde g} - \mathrm{Scal}_{\tilde g}$$
is the conformal Laplacian. The additive term is the scalar curvature of $\tilde g$.

In this case, the fact that $\tilde g$ cannot be conformal to the Euclidean metric is even more apparent, as $L_{\tilde g}$ is conformally invariant.