Let $f:\mathbb{C}\to\mathbb{C}$ be meromorphic or even entire. Let $z:\mathbb{C}\to\mathbb{C}$ be such that it holds
$$z(t+1) = f(z(t)).$$
If $f$ is entire and we choose carefully $z$ can be constructed to be also entire. This is done usually by choosing an unstable direction of a fixed point and then propagate analyticity using $f$.


So my question is the following: Given $f$ can we find $g$ such that
$$z(t+i) = g(z(t))?$$
Is this something known?
Does this $g$ depend on $z$? Is it unique?
What can we say about $g$ without knowing $z$ explicitly?


The obvious property of $g$ is that it "commutes" with $f$, i.e. $f\circ g = g\circ f$. What else can we say?


One way to look at the function $z$ is as a way to conjugate $f$ to the map $t\mapsto t+1$. So $g$ is conjugated to $t\mapsto t+i$ which is orthogonal to $t\mapsto t+1$, hence the name.


I am interested in this because it changes fixed points that are centres to nodes.


For a trivial example let's consider the map $f(z)=\frac{z}{1-z}$. The function $\Phi(z)=-z^{-1}$ conjugates it to a translation, i.e.
$$ \Phi\circ f\circ\Phi(z) = z+1. $$
Let $T(z)=z+i$, then
$$ \Phi\circ T\circ\Phi(z) = \frac{i z}{i+z}. $$



Finding the "orthogonal" map of a given 1d map
If $f$ is rational, then $g$ is also rational. It is a very rare, exceptional
situation when such two functional equations are satisfied. All such cases have been explicitly described by J. Ritt in Permutable rational functions, Trans. Amer. Math. Soc. 25 (1923), no. 3, 399–448.


If $f$ is entire transcendental, the situation is unclear. 
No single example (of such pair $f$, $g$) is known to me. Under some strong a priori conditions on $f$, I. N. Baker proved that this is impossible.


