If you didn't already know that $e^x$ is a fixed point of the derivative operator, could you still show that some fixed point would have to exist? Let's suppose you independently discovered the operator $\frac{d}{dx}$ and know only its basic properties (say, the fact it's a linear operator, how it works on polynomials, etc.) If you didn't know that $e^x$ was a fixed-point of this operator, would there be any way to (possibly nonconstructively) show that such a fixed point would have to exist? I'm curious because just given the definition of a derivative it doesn't at all seem obvious to me that there would be some function that's it's own derivative.
Note that I'm not asking for a proof that $\frac{d}{dx} e^x = e^x$, but rather a line of reasoning providing some intuition for why $\frac{d}{dx}$ has a fixed-point at all.
(And let's exclude the trivial fixed point of 0, since that follows purely from linearity rather than any special properties of derivatives).

See the wikipedia page on the theorem of Picard-Lindelöf. Your question follows by setting $f(t,y(t)) := y(t)$. The essential ingredients in the proof of this theorem are the fundamental theorem of calculus, the Banach fixed-point theorem and the fact that the space of continous functions on a compact interval equipped with the supremum norm is a Banach space. For global uniqueness of the solution you might also need Gronwall's inequality which features the exponential function, however if you are only interested in proving existence and local uniqueness you don't need Gronwall.
