Does it mean anything if the commutator of an operator with the Hamiltonian is equal to the Hamiltonian? Question says it all, really. I have $[\hat{H},\hat{O}]=-2i\hbar\hat{H}$. Does this mean that the operator $\hat{O}$ (an observable) is special in some way?
 A: $\newcommand{\ket}[1]{\left| #1 \right>}$
$\newcommand{bra}[1]{\left< #1 \right|}$
$\newcommand{bk}[2]{\left< #1 | #2 \right>}$
$\newcommand{bok}[3]{\left< #1| #2 |#3\right>}$
It means basically that all of the energy eigenstates has zero energy eigenvalue. Ups...
Let $\left| \psi \right>$ be a normalized energy eigenstate with energy eigenvalue $E_\psi$.
$$\bok{\psi}{[H,O]}{\psi}=\bok{\psi}{HO-OH}{\psi}=E_\psi \left\{ \bok{\psi}{O}{\psi} - \bok{\psi}{O}{\psi}\right\}=0$$
On the other hand:
$$\bok{\psi}{[H,O]}{\psi} = \bok{\psi}{2i\hbar H}{\psi} = 2i\hbar E_\psi = 0 \quad \forall \left| \psi \right>$$
$$\implies E_\psi=0 \quad \forall \left| \psi \right>$$
A: Suppose the space-time group includes dilatations which expand or contract space. Points in space $x^{i}\in V_{3}$ transform under a small dilatation $\epsilon$ near the identity as,
\begin{equation}
x'^{i}=x^{i}+\epsilon x^{i} \ .
\end{equation}
The change in the coords is,
\begin{equation}
\frac{d x^{i}}{d\epsilon}=x^{i}
\end{equation}
In the Hamiltonian formulation, the generator of dilatations will be some phase-space function $O$ such that the PB,
\begin{equation}
\frac{dx^{i}}{d\epsilon}=[x^{i},O]_{PB}=\frac{\partial x^{i}}{\partial x^{k}}\frac{\partial O}{\partial p^{k}}-\frac{\partial x^{i}}{\partial p^{k}}\frac{\partial O}{\partial x^{k}}=\frac{\partial O}{\partial p^{i}}=x^{i}
\end{equation}
Integrating, gives the phase space function as,
\begin{equation}
O=p^{i}x^{i} \ .
\end{equation}
If the spacetime group is Galilean relativity, the Hamiltonian is,
\begin{equation}
H=\frac{p^{i}p^{i}}{2m} \ .
\end{equation}
The PB of interest is then,
\begin{equation}
[H,O]_{PB}=-\frac{\partial H}{\partial p^{k}}\frac{\partial O}{\partial x^{k}}=-\frac{p^{k}p^{k}}{m}=-2H \ .
\end{equation}
Now go over to quantum mechanics by replacing the phase space functions with operators,
\begin{equation}
[\hat{H},\hat{O}]=-2i\hat{H}
\end{equation}
This recovers the commutator in the question and shows it has the meaning of a dilatation of Galilean space-time. 
The other answers claim that $\hat{O}$ is not Hermitian or that it does not exist. However, $\hat{O}$ must exist and be Hermitian because it's the generator of dilatations in the affine space-time and all affine spaces - those with a notion of parallelism - have dilatations in addition to translations (see chapter 13 of Coxeter's "Introduction to Geometry"). The dilatations are unfamiliar, but one can set up a similar commutator for the boost $\hat{K}$ and the arguments in the other answers would run again and say the boosts are not Hermitian or don't exist. So, the algebra for a boost is,
\begin{equation}
[\hat{K},\hat{P}]=i\hat{H}
\end{equation} 
\begin{equation}
[\hat{K},\hat{H}]=i\hat{P}
\end{equation}
Subtracting,
\begin{equation}
[\hat{P}-\hat{H},\hat{K}]=i(\hat{P}-\hat{H}) \ .
\end{equation}
This is the same as $[\hat{H},\hat{O}]=-2i\hat{H}$, modulo a numerical factor, with $\hat{H}\rightarrow\hat{P}-\hat{H}$ and $\hat{O}\rightarrow \hat{K}$.
A: 
Does this mean that the operator $\hat O$ (an observable) is special in some
  way?

I believe it means there is no such $\hat O$.
If $\hat O$ corresponds to an observable, we require the eigenvalues to be real.
Let $|o\rangle$ be an eigenket of $\hat O$ with real eigenvalue $o$:
$$\hat O |o\rangle = o |o\rangle$$
Now consider the following
$$\hat O \hat H  |o\rangle = (\hat H \hat O - [\hat H, \hat O])|o\rangle = \hat H o|o\rangle + 2i\hbar \hat H |o\rangle = (o + 2i\hbar)\hat H |o\rangle$$
Thus, $\hat H |o\rangle$ is an eigenket of $\hat O$ with complex eigenvalue $(o + 2i\hbar)$ in contradiction with the requirement that the eigenvalues of $\hat O$ are real.
