The increasing order of oxidation number of iodine in the following compounds is :

The oxidation number of H in its compound is always +1 except in metal hydrides. Therefore in HI iodine has an oxidation state of -1.

The oxidation number of an atom in its elementary form is always zero. Therefore iodine has an oxidation state of 0 in ${\mathrm{I}}_2$.

Chlorine typically has an oxidation number of -1. Considering that there is only one chlorine atom, the oxidation state of iodine in ICI would be +1.

Hydrogen is usually +1, and oxygen is -2 in most compounds. In ${\mathrm{HIO}}_4$, there are four oxygen atoms. Considering the overall charge of -1, the oxidation state of iodine would be +7.

$-1<0<+1<+7$

Therefore the correct order is $\mathrm{HI}<{\mathrm{I}}_2<\mathrm{ICI}<{\mathrm{HIO}}_4$.