We know that a reaction could have different rates at equilibrium. For example, lets take the reaction:
$$\ce{A + B -> C }$$
At equilibrium, the rates of formation and dissociation will be equal:
$$k_\mathrm{r}\mathrm{[A][B]} = k_\mathrm{c}\mathrm{[C]}$$
But that means at various concentrations of $\mathrm{[A],[B]}$ and $\mathrm{[C]}$, the actual velocity (rate) of the reaction at equilibrium will be different.
In other words, for the above example: If I were to put additional [A] into the reaction after its reached equilibrium once, the new equilibrium it would reach would have a faster rate. Also, this means that a fold change increase in $\mathrm{[A]}$ doesn't require an equivalent fold change decrease in $\mathrm{[B]}$ to reach the new equilibrium.
Now, the autoionization of water is unique because it is independent of the concentration of water: $$K_\mathrm{w} = \ce{[H+][OH-]}$$
This means that a fold change increase in $\ce{[H+]}$ requires an equivalent fold change decrease in $\ce{[OH-]}$
Thus, I am guessing the rate would be equivalent to some constant: $$k_\mathrm{r}\ce{[H+][OH-]} = constant$$
Does the rate of auto-ionization of water change when you add a strong acid or base? An explanation at the level of molecules would be appreciated.