Using the method of images, you can calculate the force between the ring of charge and the sphere.
Assume the sphere is on the z axis with it's center on the point $z$, a radius of $R_s$ and the ring's radius is $R_r$ with a charge density $\lambda$. So $z$ denotes the center of the sphere.
To calculate the force, you can replace the sphere with a charged ring (method of images) with a charge density $\lambda'$ placed at a distance $d$ below the sphere's center and a point charge $Q'$ at the center of the sphere to make the sphere electrically neutral:
$$\lambda'=-\lambda\frac{ \sqrt{R_r^2+z^2}}{R_s}$$
$$d=\frac{zR_s^2}{z^2+R_r^2}$$
$$Q'=Q\frac{R_s}{\sqrt{z^2+R_r^2}}$$
Now the problem reduces to finding the force between the main ring and the induced ring and point charge.
The image below shows the choose of parameters:
I assume that $R_{\text{ring}}>R_{\text{sphere}}$. The equations for the case of $R_{\text{ring}}<R_{\text{sphere}}$ are the same, except that the motion is restricted to $z>\sqrt{R_s^2-R_r^2}$.
The field of the main ring differs for points with $r>R_r$ and $r<R_r$:($Q$ is the ring's total charge)
$$\Phi(r,\theta)=\frac{Q}{4\pi\epsilon_0}\sum_{l=0}^\infty \mathrm{P}_{2l}(\cos \theta)\cases{\frac{R_r^{2l}}{r^{2l+1}}\mathrm{P}_{2l}(0)\,\,\,\,\,\,\,\,\,\,\text{$r>R_r$}\\\frac{r^{2l}}{R_r^{2l+1}}\mathrm{P}_{2l}(0)\,\,\,\,\,\,\,\,\,\,\text{$r<R_r$}}$$
and
$$\mathbf{E}=-\frac{\partial \Phi}{\partial r}\hat r-\frac{1}{r}\frac{\partial \Phi}{\partial \theta}\hat \theta$$
Because of the rotational symmetry of the induced charges, on which we want to find the force, the force will be in the $z$ direction: (I omitted the lengthy calculations)
$$\mathbf{F}_{\text{total}}=\hat z \frac{Q^2}{4\pi \epsilon_0} \left[\frac{z}{(R_r^2+z^2)^2}-\frac{1}{R_r^2\sqrt{z^2+R_r^2}}\cases{{\times f_{\text{out}}(r,\theta)}\\{\times f_{\text{in}}(r,\theta)}} \right]$$
Where $f_{\text{out}}$ is used when $r>R_r$ and $f_{\text{in}}$ is used when $r<R_r$. These are dimensionless functions that appear when differentiating the above potential:
$$f_{\text{out}}=\sum_{l=0}^\infty\mathrm{P}_{2l}(0)\left(\frac{R_r}{r}\right)^{2l+2}\left( (2l+1)\mathrm{P}_{2l}(\cos \theta)\cos\theta +\mathrm {P}^1_{2l}(\cos \theta)\sin\theta \right)$$
$$f_{\text{in}}=\sum_{l=0}^\infty-\mathrm{P}_{2l}(0)\left(\frac{r}{R_r}\right)^{2l-1}\left(2l\mathrm{P}_{2l}(\cos\theta)\cos \theta-\mathrm{P}^1_{2l}(\cos \theta) \sin \theta \right)$$
and $r$ and $\theta$ are functions of $z$:
$$r=\left| z-\frac{R_s^2}{\sqrt{z^2+R_r^2}} \right|$$
$$\cos \theta=\frac{z-\frac{zR_s^2}{z^2+R_r^2}}{r}$$
$$\sin \theta=\frac{\frac{R_rR_s^2}{z^2+R_r^2}}{r}$$
$\mathrm {P}^1_{2l}(\cos \theta)$ s are associated Legendre functions, and appear when differentiating $\mathrm {P}_{2l}(\cos \theta)$ w.r.t. $\theta$.
The reason of summing over even terms (indexes are $2l$ instead of $l$ ) is that $\mathrm {P}_{l}(0)$ is nonzero for even terms, as is equal to:
$$\mathrm {P}_{l}(0)= \frac 1 {2^l} \sum_{k=0}^l {l\choose k}^2 (-1)^{l-k}$$
It seems impossible to say anything intuitively, but the above equations can be used to simulate the system (with a rather good accuracy, Since the higher order terms in the above series go as $r^{2l-1}$ instead of $r^l$).
