Is there a Riemannian metric $tilde g$ on $mathbb R^d$ such that

$$tag{1}

Delta_{tilde g}=e^f(Delta +1),$$

for some $fin C^infty(mathbb R^d)$? Here $Delta=partial_{x_1}^2+ldots+partial_{x_d}^2.$

If there is such a $tilde g$, it cannot be conformal to the standard Euclidean metric $g=delta_{ij}$. Indeed, if $tilde g = e^{2phi}g$, then

$$Delta_{tilde g} = e^{-2phi} left(Delta + (d-2)g^{ij}frac{partial phi}{partial x_j}frac{partial}{partial x_i}right),$$

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.