Does ammonium react with water in an ammonia solution?
Is it possible for the ammonium to simultaneously establish another reaction with water?


It might help to add these two reactions to see what the net effect is:

$$\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-}$$
$$\ce{NH4+ + H2O &lt;=&gt; NH3 + H3O+}$$
$$+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ $$
$$\ce{2H2O &lt;=&gt; H3O+ + OH-}$$

So yes, this is the autodissociation of water, and this reaction goes on at the same time. The three equilibrium constants are linked (K1 x K2 = Kw), so there is never a situation where two of these reactions are at equilibrium and the third one isn't, causing an endless change in concentrations. In fact, for reactions in solution, there is always a state where all of the reactions are at equilibrium, and if all reactions proceed, they will all reach equilibrium eventually.

You can take any reaction that includes hydronium (or hydroxide) ions and formulate another reaction that includes hydroxide (or hydronium) ions instead by combining it with the autodissociation reaction.


  If so, we would enter a cycle where the additional concentration of ammonia resulting from the second reaction would, in turn, shift the first equilibrium to the right according to Le Châtelier's principle. 


There is no cycle, unless you call reactants making products and products making reactants at the same time a cycle. The latter is a feature of all reactions that are at equilibrium. If you label a hydrogen, it will forever cycle between being part of ammonia, ammonium, hydroxide, water, and hydronium. There are some more complicated systems of reactions where the path toward equilibrium involves oscillations of concentrations, but in most cases, concentrations approach the equilibrium concentration in a monotonic way.