Let $\Delta$ be the usual Laplacian on $\mathbb R^n$, $\Delta=\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}$. Consider the heat operator $H_t=e^{t\Delta}$. Is there an eigenfunction of $H_t$ which is not an eigenfunction of $\Delta$?
Neither operator has an eigenfunction in $L^2({\mathbb R}^n)$. But if you replace ${\mathbb R}^n$ by a bounded domain $\Omega$ with a smooth boundary, you may consider the Heat equation with the Dirichlet boundary condition $u=0$ on $\partial\Omega$. Then $e^{\Delta}$ and $\Delta^{1}$ are compact and selfadjoint on $L^2(\Omega)$, thus can be diagonalized. In addition $e^{\Delta}$ is a contraction in $L^2(\Omega)$. To see that every eigenfunction of $e^{t\Delta}$ is an eigenfunction of $\Delta$, you may use the formula $$t\Delta=\sum_{m=1}^\infty\frac1m(Ie^{t\Delta})^m.$$ This is valid over the domain $D(\Delta)$. If $e^{t\Delta}u=\lambda u$, then you obtain $$t\Delta u=\sum_{m=0}^\infty\frac1m(1\lambda)^mu=t\left(\log\frac1\lambda\right) u.$$ 

