$i\uparrow\uparrow t$
Setup
All of the relevant patterns here seem to involve only nonnegative inputs, so define $f:[0,\infty)\to \mathbb C$ by $f\left(t\right)=\begin{cases}i^{t} & \text{ if }t\in[0,1)\\i^{f\left(t-1\right)} & \text{ if }t>1\end{cases}$. To get a sense for the graph of $f$, we can use different colors for $[0,1)$, $[1,2)$,...
Graph
Here is a graph of $f(t)$ for $t\in[0,9)$:
It starts at $1$, then follows an indigo quarter-circle to $i$, then an orange sort of vertical sigmoid to $e^-\pi/2$, and then a green sigmoid up towards what appears to be a point on the initial quarter-circle, and spiraling inwards with colors like red, purple, brown, blue, yellow, and then pink.
For notational convenience, set $p=\dfrac{\pi}{2}$ and $g\left(t\right)=i^{t}=\left(e^{ip}\right)^{t}=e^{ipt}=\exp\left(ipt\right)$. This makes $f\left(t\right)=\begin{cases}g\left(t\right) & \text{ if }t\in[0,1)\\g\left(f\left(t-1\right)\right) & \text{ if }t>1\end{cases}$.
Do these actually connect? Yes
Note that $f$ is continuous (i.e. the graph is connected) since $1=g\left(0\right)$. For example, $f\left(2\right)=g\left(g\left(g\left(2-2\right)\right)\right)=g\left(g\left(2-1\right)\right)={\displaystyle \lim_{t\to2^{-}}}f\left(t\right)$. Therefore, there no breaks in the graph at $i$, $e^{-p}$, etc.
There is another sequence of apparent connections "in the middle", with the first being where the third arc (green) meets the original quarter-circle (indigo) at $\exp\left(ipe^{-p}\right)$. For this particular one, note that
$f\left(e^{-p}\right)=g\left(e^{-p}\right)=g\left(g\left(i\right)\right)=g\left(g\left(f\left(1\right)\right)\right)=f\left(3\right)$. The others are simply the result of applying $g$ to both sides: For example, since $f\left(3-\varepsilon\right)\approx f\left(e^{-p}\right)$ and there's an intersection around $f\left(4-\varepsilon\right)=g\left(f\left(3-\varepsilon\right)\right)\approx g\left(f\left(e^{-p}\right)\right)=f\left(1+e^{-p}\right)$ as well, etc.
"the branches connect perpendicularly" True.
Now we will show that the intersections in the diagram that look perpendicular all really are. To start, let's look at two particular ones.
Intersection at $i$
On $\left[0,1\right]$, we have $f\left(t\right)=g\left(t\right)$, which draws a quarter-circle in the complex plane, and it's horizontal at $i$ ($t=1$). On $\left[1,2\right]$, we have $f\left(t\right)=g\left(g\left(t-1\right)\right)$. Taking the derivative of this and evaluating at $t=1$ (to take the limit of the derivative of $f$ as $t$ approaches $1$ from above), we get $\boxed{-ip^{2}}$, so $f$ is moving vertically for $t$ just above $1$. As vertical is perpendicular to horizontal, we do have a right angle at $i$.
Intersection at $\exp\left(ipe^{-p}\right)$
Note that $f\left(e^{-p}\right)=g\left(e^{-p}\right)=g\left(g\left(i\right)\right)=g\left(g\left(f\left(1\right)\right)\right)=f\left(3\right)$. The derivative of $g\left(g\left(g\left(t-2\right)\right)\right)$ as $t$ approaches $3$ is $p^{3}e^{-p}\sin\left(p\left(1+e^{-p}\right)\right)-i\left(p^{3}e^{-p}\cos\left(p\left(1+e^{-p}\right)\right)\right)$. Since $\cos\left(p+x\right)=-\sin x$ and $\sin\left(p+x\right)=\cos x$, this simplifies to $\boxed{p^{3}e^{-p}\left(\cos\left(pe^{-p}\right)+i\sin\left(pe^{-p}\right)\right)}$.
And the derivative of $f\left(t\right)=g\left(t\right)$ at $t=e^{-p}$ is $-p\sin\left(pe^{-p}\right)+i\left(p\cos\left(pe^{-p}\right)\right)=\boxed{p\left(-\sin\left(pe^{-p}\right)+i\cos\left(pe^{-p}\right)\right)}$.
These two complex numbers are perpendicular as vectors, so this is a right angle as well.
The other intersections
To get all of the other intersections, note that $g\left(t\right)$ has a complex derivative of $ipe^{ipt}\ne0$, so it is conformal (see, for example, this MSE question), meaning that $g$ preserves the local angles in the diagram. Because of the recursive definition of $f$, this means that the two intersections considered above propagate along the diagram. For example, since there's a perpendicular intersection around $f\left(3-\varepsilon\right)\approx f\left(e^{-p}\right)$, there's a perpendicular intersection around $f\left(4-\varepsilon\right)\approx f\left(1+e^{-p}\right)$.
"Are these shapes similar to one another?" No.
I'm not certain how to interpret this question, but everything that comes to mind has the answer "no". For example, the first part of the graph of f is a quarter circle, but the fourth part (red) is certainly not.
$z\uparrow\uparrow t$
I think these questions about $z\uparrow\uparrow t$, which were added significantly after the original posting, deserve their own separate Math(s) StackExchange question. That said, here are some observations that are a bit too long for a comment.
When $|z|=1$, $\theta_0=\pi/2$? Probably.
Suppose we started with $g\left(t\right)=\exp\left(i\theta t\right)$ instead of $g\left(t\right)=\exp\left(i\frac{\pi}{2}t\right)$? Then $f\left(3\right)=g\left(g\left(g\left(1\right)\right)\right)$ is at $\exp\left(i\theta e^{i\theta e^{i\theta}}\right)$ $=\exp\left(i\theta e^{i\theta\left(\cos\theta+i\sin\theta\right)}\right)$ $=\exp\left(i\theta e^{i\theta\cos\theta-\theta\sin\theta}\right)$ $=\exp\left(i\theta e^{-\theta\sin\theta}e^{i\theta\cos\theta}\right)$ $=\exp\left(i\theta e^{-\theta\sin\theta}\left(\cos\left(\theta\cos\theta\right)+i\sin\left(\theta\cos\theta\right)\right)\right)$ $=\exp\left(\theta e^{-\theta\sin\theta}\left(i\cos\left(\theta\cos\theta\right)-\sin\left(\theta\cos\theta\right)\right)\right)$. So in absolute value, this is $\exp\left(-\theta e^{-\theta\sin\theta}\sin\left(\theta\cos\theta\right)\right)$. To determine if $f\left(3\right)$ is in the unit disc, we need to check that $h\left(\theta\right)=-\theta e^{-\theta\sin\theta}\sin\left(\theta\cos\theta\right)\le0$. Certainly $h\left(0\right)=h\left(\frac{\pi}{2}\right)=0$, and $h$ is negative in between, so when $\theta$ is slightly more than $\frac{\pi}{2}$, $f\left(3\right)$ leaves the unit disc. The next greatest zero of $h$ is at $\frac{3\pi}{2}$ (and there is another at approximately $5.34$).
However, a similar analysis shows that $f\left(2\right)$ leaves the unit disc for $\theta\in\left(\pi,2\pi\right)$ (the corresponding $h\left(\theta\right)$ is just $-\theta\sin\theta$). Therefore, $f\left(2\right)$ and $f\left(3\right)$ are both in the unit disc (considering $\theta\in[0,2\pi)$) only for $\theta\in\left[0,\frac{\pi}{2}\right]$.
I would consider this evidence suggestive that that would be the "stable range". If you were interested in $\theta\in\left[-\pi,\pi\right]$ instead, note that both $h$'s are even functions, and so we get a conjectured "stable range" of $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, which seems to agree with numerical experiment.
