Eigenfunctions of the Runge-Lenz vector The hamiltonian for the hydrogen atom,
$$
H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r}
$$
is spherically symmetric and it therefore commutes with the angular momentum $\mathbf{L}$; this causes all its eigenfunctions with equal angular momentum number $l$ but arbitrary magnetic quantum number $m$ to be degenerate in energy. 
The hydrogen atom also has a further degeneracy, in that given any angular momentum there are usually other $l$s with the same energy. This degeneracy is due to the existence of a second constant of motion, usually called the Laplace-Runge-Lenz vector,
$$
\mathbf{A} = \frac{1}{2m} ( \mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p}) - k \frac{\mathbf{r}}{r},
$$
which is the generator of an even bigger symmetry, which is isomorphic for bound states to the group $\rm{SO}(4)$ of rotations in four dimensions, of the Kepler problem.
The Runge-Lenz vector also has a rich geometrical interpretation. For a classical elliptical orbit, it points from the focus to the periapsis and its magnitude is proportional to the orbit's eccentricity. For circular orbits, it vanishes.

Image source: Wikipedia
The hydrogen atom is usually described in the common eigenbasis of the hamiltonian and the angular momentum, with the well-known and well-loved quantum numbers $|nlm\rangle$. However, the Runge-Lenz vector $\mathbf{A}$ is also a constant of the motion.
What do its eigenfunctions look like?
More concretely, I'm looking for the spatial structure of the common eigenfunctions of $H$ and at least one component of $\mathbf A$, and possibly also of $A^2$ (which, in analogy with the common eigenfunctions of $H$, $L^2$ and $L_z$, is the most one could expect), and if that's not possible then an explanation of why, and a description of suitable third quantum numbers to complete a CSCO. I would like to know what their corresponding eigenvalues are, and what the uncertainty of the other components is, whether one can assign a classical eccentricity to the orbital, and more generally in the relation to the corresponding classical geometry.
 A: Your comment inspired me to do some plotting in Mathematica :)
Following the discussion in Chapter 14 of Robert Gilmore's "Lie Groups, Physics and Geometry", the eigenstates are given by states $SU(2) \times SU(2)$ states with quantum numbers $|j_1 , m_1 ; j_2 , m_2\rangle$ with the further condition $j_1 = j_2$. (Note that the radial part is irrelevant for the discussion). The generators of the two $SU(2)$:s are given by $\mathbf{J}_1 = \frac{1}{2}(\mathbf{L} + \mathbf{A}')$ and $\mathbf{J}_2 = \frac{1}{2}(\mathbf{L} - \mathbf{A}')$, where 
$$\mathbf{A}' = \sqrt{-\frac{m}{2H}} \mathbf{A}$$
The state $|j_1 , m_1 ; j_2 , m_2\rangle$ then has eigenvalues $j_1(j_1+1)$ and $j_2(j_2+1)$ under $\mathbf{J}_1^2$ and $\mathbf{J}_2^2$, respectively, while $m_1$ and $m_2$ are the eigenvalues under the $z$-components of $\mathbf{J}_1$ and $\mathbf{J}_2$.
It is straightforward to decompose such a state in terms of eigenstates of the diagonal $SU(2)$ symmetry corresponding to angular momentum, with the coefficients being the standard Clebsch-Gordan coefficients. It is pretty tedius to do this by hand, but thankfully it can be fully automated by using Mathematica. The following code constructs the eigenstate $|J,M_1;J,M_2\rangle$ in terms of spherical harmonics (the warning it sometimes produces are hopefully nothing serious).
eigenstate[J_, M1_, M2_] := Sum[ClebschGordan[{J, M1}, {J, M2}, {j, M1 + M2}]
    SphericalHarmonicY[j, M1 + M2, \[Theta], \[Phi]], {j, 0, 2 J}]

We now just need to plot it using, eg, the code
SphericalPlot3D[Abs[eigenstate[1, 1, -1]]^2, {\[Theta], 0, \[Pi]}, {\[Phi], 0, 2 \[Pi]}, 
    PlotRange -> {{-0.25, 0.25}, {-0.25, 0.25}, {-0.7, 0.7}}]

This produces the following plot for the probability distribution in the state $|1,1;1,-1\rangle$:

Here are the states $|1/2,-1/2;1/2,1/2\rangle$ $|1,1;1,0\rangle$, $|1,0;1,0\rangle$ and $|1,0;1,1\rangle$:




The fourth state is the reflection of the second one. Note that the axes on the first one are a bit different than for the other three.
While it is easy to make more plots, it is more fun to play around with them in Mathematica where you get to rotate them and easier can see the details.
A: I think it doesn't make sense to ask for eigenfunctions of the Runge-Lenz vector. The reason is that the commutator of two components of the Runge-Lenz vector, $[A_x,A_y]$ is proportional to  $L_z$. The best that one can do is require it to be an eigenfunction of any one of the components, say $A_z$ along with the two Casimirs $J_1^2$ and $J_2^2$. So Olof's answer above is the best one can do. It is a straightforward, possibly tedious, exercise to compute the expectation value $\langle A_x\rangle$ in any of these states. The generalized uncertainty principle (aka Robertson-Schrodinger relation) can be used to estimate $\Delta A_x \Delta A_y$ in any state.
