Chemistry - Molecular structure of iodine nonoxide
Regarding its ionic form, I don' think it is as simple as $\ce{I^3+(IO3^-)3}$. It is much more complex than that. From Ref.1:
This compound is likewise to be considered as an $\ce{I(III,V)}$ oxide and reacts with alkali hydroxide to give $\ce{I-}$ and $\ce{IO3-}$.
$$\ce{3I4O9 + 12OH- -> I- + 11IO3- + 6H2O}$$
Structurally, $\ce{I4O9}$ is possibly an iodate $\ce{I3O6+IO3-}$ (more precisely $\ce{(I3O6+)_n.nIO3-}$) in which the isopolycation $\ce{I3O6+}$ has a polymeric structure and is formulated as $\ce{I_^{III}(I^{V}O3)2^+}$ consisting of twice as many pyramidal $\ce{I^{V}O3}$ groups as square-planar $\ce{I^{III}O4}$ groups thus $\ce{I4O9}$ would correspond to $\ce{I(IO3)2^+IO3-}$ which would become $\ce{I(IO3)3}$.
Reference
- Inorganic Chemistry By Egon Wiberg, A. F. Holleman, Nils Wiberg, Academic Press, 2001