I like this question because it really works your intuition.
The basics: x^y means "grow at x, for y units of time". I see "2^3" as "grow at 2x for 3 units of time".
Having a base of i means your "growth" is a rotation at 90 degrees, no scaling. So i^(1/2) means a 45 degree rotation, i^3 means a 270 rotation, etc.
Raising this to the i power (or 1/i power, which is -i) means the growth that was originally purely rotational is now rotated. So instead of growing at i, you are growing at (i * 1/i = 1). So, we should expect a positive real number, greater than 1 since our growth is positive.
How long do we actually grow for? Well, the base of "i" is really e^(i* pi/2), which means "Start at 1.0 and rotate continuously for pi/2 seconds". We've now modified this to "Start at 1.0 and grow at 1.0 for pi/2 seconds", which is e^(pi/2).
So the answer is i^(1/i) = e^(pi/2) ~ 4.8
It's a bit tough with text-only, read the above article for more diagrams.
It is countably infinite. Let's try to find them all.
A relatively simple way to understand this is that i^i = e^(i * log(i)) for every possible log of i. So all we need to do is understand what values log(i) could have (there are actually many), and then we can work it out. But log(z) just undoes e^z, so we need to understand e^z.
Now let's work backwards. If z = x + y i with x and y real, then x tells us the absolute value of e^z and y tells us the angle. The absolute value of i is 1, so any possible solution to log(i) has real part 0. The angle that we want to wind up with is 90 degrees, or pi/2. Therefore y can be ..., -3.5 pi, -1.5 pi, .5 pi, 2.5 pi, 4.5 pi, ... .
Therefore log(i) has to be one of 1.5 pi i, -.5 pi i, -2.5 pi i, -4.5 pi i, ... .
Now i^i is e^(i log(i)) so it can be any of ..., e^(3.5 pi), e^(1.5 pi), e^(-.5 pi), e^(-2.5 pi), e^(-4.5 pi), ... .
Unless I've made a trivial calculation error, that is the whole list.
This is of course much trickier because to define z^w for complex numbers requires that one choose a complex logarithmic function. There are infinitely many choices for this. Everyone has decided to use the principle logarithm but it could be consistently defined using any branch cut.
To answer the question, e^(it) =cos(t)+ i * sin(t). That means that when t=pi/2, cos(t)=0 and sin(t)=1, so i = e^(i * pi/2)
To get the 'i'th root we divide the exponent by i, so i^(1/i) = e^((i * pi/2)/i) which is e^(pi/2).
As for the angle metaphor, that works for multiplying by complex numbers. The usual "understanding" of exponentiation is repeated multiplication, but taking a power of "i" can't be understood as multiplying together "i" lots of the number. Similarly the "i"th root can't be understood like that.
This is is because exponentiation to an imaginary power can also be thought of as rotation. You have presumably heard e^(pii) = -1. That's because it's a half-rotation away from 1. Half of that half rotation would be i, and is sqrt(e^(pi i)), which is e^(pi/2*i), which is equivalent to (e^(pi/2))^i, and the ith root of that is obviously (e^(pi/2)).
