How can I intuitively understand complex exponents?
In brief: thinking of exponentiation as repeated multiplication is equivalent to the identity $a^{m+n}=a^ma^n$. But focusing on the latter equation instead of the former concept allows us to expand exponentiation beyond the concept of "multiply the base $n$ times", to negative, rational, real, and complex exponents. As you say in the OP question, "the basic rules" allows you to extend past the notion of repeated multiplication for counting numbers. Focusing instead on the equation $a^{m+n}=a^ma^n$ means following the mantra: exponentiation turns additions into multiplications, and rotations are a natural multiplication on the complex plane.
In more detail:
The notion of exponentiation as repeated multiplication is, for natural numbers $n,m$ equivalent to the identity $a^{m+n}=a^ma^n$, because
$$a^{n} = a^{\underbrace{1+\dotsb+1}_{n\text{ times}}}=\underbrace{a\cdot\dotsb\cdot a}_{n\text{ times}}.$$
By focusing on this identity moreso than the notion of "multiplication repeated $n$ times", it also allows us to make sense of exponentiation of naturals, integers, rationals, reals, complexes, even matrices and more, whereas the repeated multiplication notion only makes sense for $n$ natural number, a counting number. We only extend to exponentiation of zero, negatives, and rationals via the above identity (or other similar). Therefore we should view the identity $a^{m+n}=a^ma^n$ not just as a consequence of exponentiation as repeated multiplication, but as a complete and fundamental conceptual replacement. As our conceptual starting point. Exponentiation is, by definition and fundamental conception, the operation that turns addition into multiplication.
As you say in your question, you must extract an understanding of exponentiation of rational, negative, and real exponents via the "basic rules". This identity $a^{m+n}=a^ma^n$ is the most basic of our basic rules, and it will guide us in extracting an understanding imaginary exponents as well.
Now to the matter at hand. On the real line, real numbers act additively by shifting, and multiplicatively by scaling away from zero.
On the complex plane, real numbers act additively by shifting along the real axis (horizontally), imaginary numbers act additively by shifting along the imaginary axis (vertically). Note that orbits of these two actions are orthogonal. Horizontal lines versus vertical.
Real numbers act multiplicatively by stretching away from the origin, while imaginary numbers by rotating $90º$. Note that the orbits of these two actions are also orthogonal. Orbits of scalings are radial lines; orbits of rotations are circles.
Now we have decided that our most fundamental identity of exponentials is $a^{x+y} = a^xa^y$. Exponentiation turns adders into multipliers. It turns real adders (i.e. horizontal shifts) into real multipliers, i.e. scalings away from zero. Horizontal lines into radial lines.
And therefore what must exponentiation transform the orthogonal imaginary shifts, i.e. vertical shifts into? They must transform into those multipliers which are orthogonal to the radial expansions. Which are the rotations. Vertical lines into circles. So exponentiation with an imaginary exponent must be a rotation.
The base of the exponentiation sets the size scale of these stretchings and rotations, and exponentiation with base $e$ does natural rotations in radians, but this picture works with any base of exponentiation, so long as $a>1.$
This intuition is encoded in Euler's identity $e^{i\theta}=\cos\theta + i\sin\theta$. A special case is $e^{i\pi} = -1$, which just says that rotation by $180º$ is the same thing as reflection. This intuitive point of view for understanding Euler's identity is explained in a popular 3Blue1Brown video.
So how do we understand an expression like $x^{\frac{i\pi}{4}}$? Well assuming $x$ is real with $x>0$, since the exponent is imaginary, it's a rotation. How big a rotation? well it depends on the base $x$, and the exponent $\frac{i\pi}{4}$. Computing this magnitude could be thought of as an exercise in operations derived from repeated multiplication, as you set out in your question, but doing so is of limited utility.
Instead we should think of it as an adder $\frac{\pi}{4}$, turned into a vertical shift $\frac{i\pi}{4}$ by the complex unit $i$, and then turned into a rotation by exponentiation with base $x$ (as long as $x$ is real and $x>1.$) The magnitude of $x$ determines the speed of the rotation, or the units.
I think it would be helpful for you to let go of the idea (at least in complex analysis) that $a^b$ means multiplying $a$ $b$ times by itself. Rather, exponentiation starts at the function $\exp : \mathbb C \to \mathbb C$, which you probably usually denote as $\exp(a) = e^a$. Then (once you have chosen a branch of the complex logarithm) you can define $a^b = \exp(\log(a) \cdot b)$. With that definition, all the repeated multiplication rules that you know come down to the identities $\exp(a + b) = \exp(a)\exp(b)$ and $\exp(0) = 1$.
Rather than trying to press complex exponentiation into the mold of repeated multiplication, see complex exponentiation -- or more fundamentally, the function $\exp$ -- as its own thing, that in special cases can be interpreted as repeated multiplication, thanks to the functional relation $\exp(a)\exp(b) = \exp(a + b)$.
I guess we agree on $ e^{a + i \phi} = e^{a} \cdot e^{i \phi} $. So let's look at the latter term with imaginary exponents.
Imaginary exponents are indeed of a different quality then real ones. Maybe your best approach is to realize that the justification for the "magic" Euler equation
$$ e^{i \phi} = \cos(\phi) + i \sin(\phi) $$ only comes from the formal continuation of the exponential function from real to complex domain. Indeed, by just taking the series expansion of the exponential function, $e^x = \sum_{n=0}^\infty x^n/n!$, apply for $x = i \phi$, and grouping real and imaginary parts you see that this holds true. This gives you (for odd and even powers of $\phi$), with real $\phi$, of course:
$$ e^{i \phi} = \sum_{n=0}^\infty (i\phi)^n/n! = \sum_{n=0}^\infty (-1)^n \phi^{2n}/(2n)! + i \sum_{n=0}^\infty (-1)^n \phi^{2n +1}/(2n+1)! $$ and the two sums are just the expansions of $\cos \phi$ and $\sin \phi$, for real $\phi$.
So yes, it is unintuitive that you now stay on the unit circle in the complex domain, that it's cyclic in $2 \pi$, etc.
EDIT: (see also Paul Sinclair's comment)
If you know about the derivatives of $\cos \phi$ and $\sin \phi$, here's an approach without series expansions.
We know that $\cos' \phi = - \sin \phi$, and $\sin' \phi =\cos \phi$. Now define a function $ f(\phi) = \cos(\phi) + i \sin(\phi) $. Then you have $ f'(\phi) = - \sin(\phi) + i \cos(\phi) = i f(\phi) $.
On the other hand, we know for real constants $a$ that the equation $ f'(\phi) = a f(\phi) $ has a unique solution, which is $f(\phi) = c \cdot e^{a \phi}$ with some constant $c$ which is determined by some initial condition. Now comes the trick again of the formal continuation of the exponential function from real to complex domain. We allow now that $a$ may as well be complex. Then we have that $ f'(\phi) = i f(\phi) $ (as taken from the trig function above) has the unique solution $f(\phi) = c \cdot e^{i \phi}$. Determine the constant $c=1$ by observing $f(0) = c \cdot e^{i 0} = c$ and also $ f(0) = \cos(0) + i \sin(0) = 1 $.
So this explains Euler's equation as well.