Why does $\ce{NH3+}$ have a higher $-\ce{I}$ effect ( inductive effect) than $\ce{NO2}$ even though oxygen is more electronegative? Does it have something to do with positive charge on $\ce{N}$?
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2 Answers
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The short answer is that $\ce{NH3+}$ is a cation, and any cation will always draw electrons towards it stronger than any neutral species.
The electronegativity of the species involved is only a secondary cause. However, one can safely say that $\ce{NH2}$, if nitrogen’s lone pair is (somehow) impeded from excersizing it’s $+\ce{M}$ effect, would have a lesser $-\ce{I}$ effect on the whatever than $\ce{NO2}$ would.
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The inductive effect which you are talking about must be from the most common order given for strength of -I.
Here what you are missing is that its not $\ce{NH3+}$ but its - $\ce{NH3+}$ or $\ce{R-NH3+}$.
What I mean to say is some alkyl or aryl group will be attached to it.
Coming to your Question, -I, means that the group attached will pull electrons more towards itself, gaining partial negative charge and hence imparting partial positive charge to the alkyl or aryl group . Though $\ce{O}$ is more EN than $\ce{N}$, $\ce{NO2}$ is less EN comparatively because there is cation on $\ce{NH3+}$, so it will have more tendency to gain or pull an electron.
