How to decrease the percent of ammonia that is converted to the ammonium ion in water?
If ammonia gas is bubbled into pure water, the following reaction takes place:

$$\ce{NH3(aq)  +   H2O(l)   &lt;=&gt;     NH4+(aq)  +    OH−(aq)}\tag{1}$$

How to decrease the percent of ammonia that is converted to the ammonium ion in water?

Le Chatelier's Principle indicates that adding $\ce{OH-}$ would push the equilibrium to the left. 

I lifted the following figure from:


  HYDROPHOBIC MEMBRANE TECHNOLOGY FOR AMMONIA EXTRACTION FROM WASTEWATERS by Airton Kunz and Saqib Mukhtar Eng. Agríc. vol.36 no.2 Jaboticabal Mar./Apr. 2016
  https://doi.org/10.1590/1809-4430-Eng.Agric.v36n2p377-386/2016  




As can be seen in the figure, an increase in pH favors formation of $\ce{NH3}$ and a decrease in the pH favors $\ce{NH4+}$.


  Since the OP asked only for the percent of $\ce{NH4+}$ to be reduced, there is another way. Dilute the solution.


Will $\ce{Fe(NO3)3}$ decrease the percent of ammonia that is converted to the ammonium ion in water?

Iron (III) forms an relatively insoluble hydroxide as shown by the following equation:

$$\ce{Fe^{3+} + 3OH- &lt;=&gt;[K_{\mathrm{sp}} = 2.79\times10^{−39}] Fe(OH)3(s)}\tag{2}$$  

So again according to Le Chatelier's Principle removing $\ce{OH-}$ from reaction (1) favors the reaction moving to the right. Thus the precipitation of $\ce{Fe(OH)3}$ will increase the formation of $\ce{NH4+}$.


  In a comment on the question user Poutnik indicated the reaction $\ce{Fe^{3+} + 3NH3 + 2H2O -&gt; FeO(OH) + 3NH4+}$.  I'm sure there is a study somewhere on $\ce{Fe(OH)3 \pu{ vs } FeO(OH)}$ formation, but I didn't find it in 15 minutes of fooling around. I'd expect that strong $\ce{OH-}$ concentrations and higher temperatures (boiling solution) would favor $\ce{FeO(OH)}$.