f(x)=1-(1/x)=(x-1)/x.
f(f(x))=(f(x)-1)/f(x)=
((x-1)/x-1)/((x-1)/x)=
((x-1-x)/x)/((x-1)/x)=
-1/(x-1).
f(-1/(x-1))=((-1/(x-1))-1)/(-1/(x-1))=
((-1-x+1)/(x-1))(-(x-1))=x.
Therefore, f(f(f(x)))=x.
[For example, if x=2, f(2)=½, f(½)=-1, f(-1)=2. So f(f(f(2)))=2=x.]
i_A may indicate a definition of i_A. But I think it means any element of set A. So the compound function maps each element onto itself. Zero is excluded because it would generate 1/0 which is undefined.