Jabotinsky's iterative logarithm

[bo198214]

Would that be true also for imaginary t? t=I, like:

\( \text{ilog}(f^{\circ I}) = I \text{ilog}(f) \)

yes. As one can see from the derivation it is true for any complex iteration counts.

The functions having ilog property ilog(fg)=ilog (f) + ilog(g) seem to be rather wide class, am I right?

Its not one class, there are equivalence classes where \( f \) is considered to be equivalent to \( g \) if there is some \( t\neq 0 \) such that \( f=g^{\circ t} \), for each equivalent \( f \) and \( g \) this property holds.