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).

Suppose you have a small positive number $\epsilon$. Consider functions which map integer multiples of $\epsilon$ onto $\mathbb{R}$. These functions have an approximate derivative 
$$f'(i\epsilon)\approx \frac{f((i+1)\epsilon)-f(i\epsilon)}{\epsilon}$$
(for integer $i$)
I think that it is intuitively clear that for these functions and this approximate derivative, the approximate derivative has a fixed point.
It can be constructed trivially as follows: define $f(0)=1$, $f((i+1)\epsilon)=f(i\epsilon)+\epsilon f'(i)$.
Of course, this approximate derivative approaches the true derivative in the limit $\epsilon\rightarrow 0$. If you expect that this line of reasoning still works "in the limit", then you expect that $\frac{d}{dx}$ has a fixed point.
