Decomposition of ammonium nitrate

Consider the decomposition of $\ce{NH4NO3 -> N2O +2 H2O}$


How do I show this is a redox reaction. Within the $\ce{H2O}$ the oxidation state of $\ce{O}$ is $-2$ and the oxidation state of $\ce{H}$ is $+1$.
As ammonium nitrate is not a peroxide the oxidation state of $\ce{O}$ should also be $-2$. Hydrogen should be $+1$ again. It seems there is no transfer of electrons hear so somehow nitrogen must be both losing and gaining electrons. How is this possible?


Furthermore is this a disproportionation reaction as nitrogen both loses and gains electrons?




As the [oxidation state/number](https://en.wikipedia.org/wiki/Oxidation_state) (ON) is generally a formal charge, oxidation/reduction is generally a formal electron exchange. Even if in many cases it is literal exchange.


Exact reaction mechanism is not probably known.


Formally, the nitrogen atom of ammonium is oxidized from $\text{ON}=-3$ to $\text{ON}=0$ by the nitrate nitrogen, which is reduced from $\text{ON}=+5$ to $\text{ON}=+2$.


Depending on chosen formal reaction steps, oxidation of ammonium and reduction of nitrate can be done by formal homolytic break of bonds with hydrogen or oxygen, where formally free hydrogen and oxygen later recombine to water.




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[Disproportionation](https://en.wikipedia.org/wiki/Disproportionation) in Redox reaction context are two simultaneous redox reactions, leading from the same oxidation number to two different ones, like $\ce{2 Cu+(aq) -> Cu^2+(aq) + Cu(s)}$.


[Comproportionation (synproportionation)](https://en.wikipedia.org/wiki/Comproportionation) is the opposite, like $\ce{SO2 + 2 H2S -> 2 H2O + 3 S}$.


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