Jabotinsky's iterative logarithm

[Ivars]

This must mean that:
\( \text{ilog}(f^{\circ t}) = t \text{ilog}(f) \)
so
\( f^{\circ t} = \text{ilog}^{-1}(t \text{ilog}(f)) \)
Andrew Robbins

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

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

\( f^{\circ I} = \text{ilog}^{-1}(I \text{ilog}(f)) \)

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

Would it be true for ( I am only writing down iterates of f):

any 1/m, 1/n - are there any constraints on m, n like they must have common integer which they both divide?

any 1/x, 1/y or just 1/x, 1-1/x = (x-1)/(x), or n/x , 1-n/x = (x-n)/x?

purely complex conjugate iterates, like it, -it ?

z and z*?

Please tell me where I am first wrong in my conjecturing.

Best regards,

Ivars