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).

You want a function $f$ such that $f'=f$. Let's hope that such function exists and has an inverse, $g$ (this is just for motivation: we will prove it in the long run).
We then have that, since $f \circ g =Id$, by the chain rule, $f'(g(x)) g'(x)=1$. Therefore,
$$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{f(g(x))}=\frac{1}{x}.$$
The derivative of this inverse seems more manageable, since it does not depend on $g$ itself (The motivation ends here). Let's then define $g:\mathbb{R}_{>0} \to \mathbb{R}$:
$$g(x)=\int_1^x\frac{1}{t}dt.$$
By the FTC, $g$ satisfies what it needs. If we prove that $g$ is invertible, we are done.
For that, first note that $g(xy)=g(x)+g(y)$. To see this, let $y$ be fixed. Then, $(g(xy))'=\frac{1}{xy}y=\frac{1}{x}$, and the derivative of the right side is also trivially $\frac{1}{x}$. Since both sides coincide in $1$ by a trivial checking, both must be equal. Since $y$ is arbitrary, the relation holds for any $x,y$ as we wanted.
Now, note that it follows that $g(2^n)=n g(2)$. But $g(2)>0$ by its very definition. Since $g$ is increasing, it follows that $g(x) \to \infty$ as $x \to \infty$. Since $g(1)=g(x)+g(\frac{1}{x})$, it follows that $g(x) \to -\infty$ as $x \to 0$, and we are finished proving that $g$ is a bijection due to the IVT.
Now, take the inverse $f$ of $g$. We will have by the chain rule that $$g'(f(x))f'(x)=1 \implies f'(x)=f(x).$$
