The origin of the question is the computation of two integrals $$F(x) = \int_0^x \frac{\mathrm{d}y}{\cos(y)} \textrm{ for } |x|<\pi/2,$$ $$\tilde F(x) = \int_0^x \frac{\mathrm{d}y}{\cosh(y)} \textrm{ for } x \in \mathbb{R}.$$ Classical formulae yield $$F(x) = \ln \tan \Big(\frac{x}{2}+\frac{\pi}{4}\Big),$$ $$\tilde{F}(x) = 2\arctan(e^x)-\frac{\pi}{2} = 2\Big(\arctan(e^x)-\frac{\pi}{4}\Big).$$ What is surprising is that $\tilde{F}$ is the inverse bijection of $F$!
At the same time, the functions $F$ and $\tilde{F}$ can be extended into holomorphic functions, at least on the simply connected open sets $\{z \in \mathbb{C} : |\Re(z)| < \pi/2\}$ and $\{z \in \mathbb{C} : |\Im(z)| < \pi/2\}$. And because of the relation $\cosh(z) = \cos(iz)$ (and vice-versa), these two functions are related by $\tilde{F}(z) = F(iz)/i$.
How one can « explain » this coincidence? Do we have other non-trivial examples?
Alternative formulas for $F$ and $\tilde{F}$ give us a hint: $$F(x) = \ln \frac{1+\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})} = 2 \arg\tanh \tan(x/2),$$ $$\tilde{F}(x) = 2\arctan\Big(\frac{e^x-1}{e^x+1}\Big) = 2\arctan \tanh(x/2).$$ Now, we see why $\tilde{F}$ is at the same time the inverse bijection of $F$ and given by $\tilde{F}(z) = F(iz)/i$.
And we know how to construct other examples. Indeed, set $H(z)=iz$. Let $G$ be any holomorphic function defined on an open set containing $0$ such that $G(0)=0$ and $G'(0) \ne 0$. Hence (by local inversion theorem for holomorphic functions) the function $G$ can be reversed on the neighborhood of $0$. Assume that the first non-zero Since $G$ and $H^2$ commute, the function $F := G \circ H \circ G^{-1} \circ H^{-1}$ satisfies $$F^{-1} = H \circ G \circ H^{-1} \circ G^{-1} = H^{-1} \circ G \circ H \circ G^{-1} = H^{-1} \circ F \circ H.$$
Question: Does this method provide all power series such that $f(z) \sim z $ as $z \to 0$ and whose local inverse (for the composition law $\circ$ ) is given by $\tilde{F}(z) = F(iz)/i$?
We get a slightly simpler variant of this question by replacing $H$ by the map $z \mapsto -z$ and by removing the assumption $G$ is odd.
Let us give an example : set $G(z) = z+z^2$, so $G^{-1}(z) = \frac12(-1+\sqrt{1+4z})$ at the neighborhood of $0$, where $\sqrt{}$ is the principal determination of the square root on $\mathbf{C} \setminus \mathbf{R_-}$. Then we get $$F(z) = G \big(-G^{-1}(-z) \big) = 1-z-\sqrt{1-4z},$$ $$F^{-1}(z) = -G \big(-G^{-1}(z) \big) = 1-z+\sqrt{1+4z}$$