Is there a metric on Euclidean space that turns the Helmholtz equation into the Laplace equation? 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.
 A: After Terry Tao's comments, I came to the conclusion that the only possible choice of a metric $\tilde g$ and of an operator
$$T_{\tilde g}= \Delta_{\tilde g} + \text{lower order terms} $$
that will give
$$\tag{1} T_{\tilde g}=e^f(\Delta +1)$$
is the following, in Cartesian coordinates:
$$\tag{2}
\tilde g_{ij}= e^{2\phi}\delta_{ij},\qquad T_{\tilde g}=\Delta_{\tilde g}-e^{-2\phi}\delta^{ij}\partial_i \phi \partial_j + e^{-2\phi}.$$
That is, $\tilde g$ must be conformal to the standard Euclidean metric, with an arbitrary conformal factor $e^{2\phi}$. I have no idea whether the operator $T_{\tilde g}$ has a geometric meaning, or if this small computation can ever be useful. In any case I am posting it here.
Proof.
We can write (1) as
$$
\tilde{g}^{ij}\partial_i\partial_j + a^k\partial_k +c= e^f \delta^{ij}\partial_i \partial_j + e^f, $$
for some scalar fields $a_k$ and $c$. This clearly implies $\tilde g^{ij}=e^f \delta^{ij}$, which is the first equation in (2), and also $a_k=0, c=e^{-2\phi}$. Since
$$\Delta_{\tilde g}= e^{-2\phi} \left(\Delta + (d-2)g^{ij}\frac{\partial \phi}{\partial x_j}\frac{\partial}{\partial x_i}\right),$$
the only possibility is to define $T_{\tilde g}$ like in the second equation in (2).
