Can one hear the shape of a drum for operators? M. Kac in his famous paper "Can one hear the shape of a drum?" asked whether one can "hear" the area of the ambient domain by looking at the spectral picture. Although he was not the first who came up with this problem, his paper had a big impact in publicizing the matter, which was proved (Gordon et al, J. Milnor etc.) to be, in general, false.
Let's pose the same question for self-adjoint compact integral operators defined on $L^2(D,dA)$ where $D$ is a bounded plane domain and $dA$ is the area measure.
More precisely: does there exist an integral operator with $L^2$ or continuous kernel so that identical spectra would lead to "similarity" of defining domain?
$\textbf{Edit 1:}$ Thanks to Ben Linowitz, there was some chronological inaccuracies, which I corrected them.
$\textbf{Edit 2:}$ The operators of interest are induced by continuous kernels that are defined everywhere.(for instance one can think of logarithmic potential for planar case, or Newtonian also Riesz  potentials for higher dimensions)
 A: This is an extended comment.
First, the chronology in your question is a bit off. Mark Kac's famous "Can you hear the shape of a drum" paper was published in 1966, two years after Milnor's examples of isospectral non-isometric 16-dimensional flat tori were published. So it was already known that the isometry class is not an invariant of the Laplace spectrum in general. What Kac's paper did was popularize this question for domains in the plane. And this is what Gordon, Webb and Wolpert proved was impossible.
On the other hand, Wolpert had proven in 1977 that in certain situations the isometry class is determined by the Laplace spectrum. More precisely, he showed that this holds for a generic compact Riemann surface.
In their proof, Gordon, Webb and Wolpert made use of an extension of what is now called Sunada's Method. This method actually produces manifolds (or orbifolds) which are strongly isospectral, hence isospectral for any natural self-adjoint elliptic operator (for instance the Laplacian acting on $p$-forms). Sunada's method is by far the most commonly used method of constructing isospectral manifolds, so there are all sorts of examples of operators, in all sorts of settings, whose spectra do not determine isometry class.
A: I think this question is ill-posed, because you cannot consider the same operator on different domains (i.e. on different $L^2(\Omega)$ spaces). Hence you would have to require a certain "functoriality" or so, and I don't know if it is clear what you would exactly want here.
Still, there are cheap answers to your question. For example, you can consider $\Delta^{-s}$, which will have a kernel as regular as you like if you make $s$ large enough, and two domains are distinguishible by its spectrum if and only if they are distinguishable by the spectrum of $\Delta$.
To construct an operator that definitely distinguishes all domains in $\mathbb{R}^n$, you can for example make the following construction. Take the indicator function of your domain $\Omega$ and consider its Fourier transform $F_{\Omega}$, which will be an entire function since $\Omega$ is compact. Let $c_\alpha$ be the coefficients of its Taylor series at zero. For any orthonormal basis $\phi_\alpha$ of $L^2(\Omega)$ (indexed by $\alpha$), the kernel
$$k(x, y) := \sum_{\alpha \in \mathbb{N}^\alpha_0} c_\alpha \varphi_\alpha(x) \varphi_\alpha(y)$$
will be $L^2$ by the Cauchy-Hadamard theorem (and I think even smooth). The associated operator has the $c_\alpha$ as eigenvalues two operators constructed with this recipe will have the same eigenvalues if and only if the two domains are equal.
