The question
Using only the principle branch of the complex log.
Skipping to the end, if $z = \frac{2i}{\pi}W(-\frac{\pi i}{2})$, which is the fixed point of $z \rightarrow i^z$, it's not true that a general complex perturbation to this value ($\epsilon$) under iteration will necessarily shrink under iteration $(z+\epsilon) \rightarrow i^{z+\epsilon} = z + \epsilon'$. So it does not converge smoothly (there's the possibility at each iteration it gets farther away). Does it converge anyway? (It seems to.)
You can see the various ways that the abs and arg of $\epsilon$ affect the abs of $\epsilon'$ at this desmos graph (in this graph $\lVert \epsilon \rVert$ is x, and $\angle \epsilon$ is t). Observe that even at very small magnitude x, the resulting abs is larger than the original more often than not: ($\theta \in [-2.62\dots, -0.53\dots] \cup [0.53\dots, 2.62\dots]$ or about 67% of the range)
From this, it makes me suspicious that it might not converge, but be oscillatory around a very small area near $z$. Can it be proven that this value is actually the convergent value? And is the approach below salvageable?
My initial approach
The approach is to use a small perturbation $\epsilon$ and prove that the result of the transform $z+\epsilon \rightarrow i^{z+\epsilon}$ is always closer to $z$ than $z+\epsilon$ is. Formally, $\lVert ϵ \rVert > \lVert i^{z+\epsilon} - z \rVert$.
Once proved, demonstrate that it works on a disc centered around $z$ that includes a value that we know is reachable from $i$, $i^i$, $i^{i^i}$, $\dots$
$$ \begin{align} i^{z+\epsilon} - z &= i^z i^\epsilon - z \tag{a} \\ &= z i^\epsilon - z \\ &= z(i^\epsilon - 1) \\ \lVert z(i^\epsilon - 1) \rVert &= \lVert z \rVert \lVert i^\epsilon - 1 \rVert \end{align} $$
For line (a) this is only true for particular branch cuts, but it should be possible to always find a branch cut that works.
Let $\epsilon = \rho \mathrm{e}^{i \theta}$ and $\epsilon' = \rho' \mathrm{e}^{i \theta'}$.
$$ \begin{split} i^\epsilon &= i^{\rho \mathrm{e}^{i \theta}} \\ &= {\left( \mathrm{e}^{\frac{\pi}{2} i} \right)}^{\rho \mathrm{e}^{i \theta}} \\ &= \mathrm{e}^{\frac{\rho \pi}{2} i \mathrm{e}^{i \theta}} \\ &= \mathrm{e}^{\frac{\rho \pi}{2} \mathrm{e}^{i \left(\frac{\pi}{2} + \theta\right)}} \\ &= \mathrm{e}^{\frac{\rho \pi}{2} \left( \cos\left(\frac{\pi}{2} + \theta\right) + i \sin\left(\frac{\pi}{2} + \theta\right) \right) } \\ &= \mathrm{e}^{\frac{\rho \pi}{2} ( -\sin(\theta) + i \cos(\theta) ) } \\ &= \mathrm{e}^{- \frac{\rho \pi}{2} \sin(\theta)} \mathrm{e}^{i \frac{\rho \pi}{2} \cos(\theta) } \\ \end{split} $$
So we have:
$$ \begin{split} \rho' &= \mathrm{e}^{- \frac{\rho \pi}{2} \sin(\theta)} \\ \theta' &= \frac{\rho \pi}{2} \cos(\theta) \end{split} $$
Next, find the abs of the iterated perturbation.
$$ \begin{split} \lVert \epsilon' - 1 \rVert &= \lVert \rho' \mathrm{e}^{i \theta'} - 1 \rVert\\ &= \lVert \rho' \cos(\theta') + \rho' i \sin(\theta') - 1 \rVert \\ &= \sqrt{{\left(\rho' \cos(\theta') - 1\right)}^2 + {\left(\rho' \sin(\theta')\right)}^2} \\ \end{split} $$
This is where the equation in the graph came from. From here, I initially attempted to simplify in the following way, but as I went to explain my result to someone else I realized that the following wasn't justified.
$$ \begin{split} &= \sqrt{\rho'^2 + 1 - 2 \rho' \cos(\theta')} \\ &= \sqrt{\mathrm{e}^{- \rho \pi \sin(\theta)} + 1 - 2 \mathrm{e}^{- \frac{\rho \pi}{2} \sin(\theta)} \cos\left(\frac{\rho \pi}{2} \cos(\theta)\right)} \\ &< \sqrt{\mathrm{e}^{- \rho \pi} + 1 - 2 \mathrm{e}^{- \frac{\rho \pi}{2}}} \\ &< \lVert \mathrm{e}^{-\frac{r \pi}{2}} - 1 \rVert \\ &< 1 - \mathrm{e}^{-\frac{r \pi}{2}} \\ &< \frac{r \pi}{2} \end{split} $$
(So finally because $\lVert z \rVert \left(\frac{\pi}{2}\right) < 1$ the value converges. Or it would if those steps were justified.)
Appendix: Derivation of the fixed point.
$$ \begin{split} z &= i^z = e^{i \frac{\pi}{2} z} \\ z e^{-i \frac{\pi}{2} z} &= 1 \\ - \frac{i \pi}{2} z e^{- \frac{i \pi}{2} z} &= - \frac{i \pi}{2} \\ - \frac{i \pi}{2} z &= W(- \frac{i \pi}{2}) \\ z &= \frac{2i}{\pi} W(- \frac{i \pi}{2}) \\ \end{split} $$ If you skipped to the end looking for the question, it's at the end of the first section.
