Solve for z.
$\displaystyle i^{z} = z$ where $\displaystyle z \in \mathbb{C} $
OK. Do this instead. Let a be any constant (real or complex).
$\displaystyle a^{z} = z $
$\displaystyle a^{-z} = \frac{1}{z} $
$\displaystyle z a^{-z} = 1 $
$\displaystyle -z a^{-z} = -1 $
$\displaystyle -ze^{-z \ \text{Log a}} = -1 $
$\displaystyle -z \ (\text{Log a}) \ e^{-z \ \text{Log a}}= - \text{Log a} $
$\displaystyle W(-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}) = W (- \text{Log a}) $
$\displaystyle -z \text{Log a} = W(-\text{Log a}) $
$\displaystyle z = \frac{W(-\text{Log a})}{-\text{Log a}} $
$\displaystyle \text{Log i} = \ln|i| + i \frac{\pi}{2} = i\frac{\pi}{2}$
$\displaystyle \frac {W(-i \frac{\pi}{2})}{-i\frac{\pi}{2}} \approx 0.4382829366+i0.3605924718$
Is that satisfactory?
It is better in that you eliminate the $\displaystyle \log$ 's in the final answer but I would still prefer a solution that did not have to worry about which branch of the $\displaystyle \log$ was in use where. That can be done by the the first step that I proposed (which, while that is not the way I think of it, absorbs the taking of logarithms while effectively fixing the branch).
Also as $\displaystyle W$ is multi-valued the right hand side of final line is just one of the solutions that correspond to the left hand side (as you know, and I know you know).
CB
I like this method:
As $\displaystyle i=e^{(2n+1/2)\pi i}=e^{-\mu_n}$ where $\displaystyle \mu_n=-(2n+1/2)\pi i$ for $\displaystyle n\in\mathbb{Z}$ we have:
$\displaystyle e^{-\mu_n z}=z$
or:
$\displaystyle \mu_n z\; e^{\mu_n z}=\mu_n$
so:
$\displaystyle \mu_n z=W(\mu_n)$.
Hence:
$\displaystyle z=\frac{1}{\mu_n}W(\mu_n); \ \ \mu_n= -(2n+1/2)\pi i,\ n \in \mathbb{Z}$
(notes:
1. this is at least the third refinement of the original idea scribbled in my note book
2. the multi-valuedness of the solution is not only dependent on $\displaystyle n$, but the Lambert-W is itself multi-valued)
CB