A function that satisfies the $n$-th derivative where $x=0$ is $\frac{1}{n}$ Is there a function that satisfies $f^{(n)}(0)=\frac{1}{n}$ for every positive integer $n$?
If you assume $\forall n\geq 1$, then I gather any primitive of $f(x) =\frac{e^x-1}{x}$ should do.

The rationale behind it is to find a power series $F$ with $F(x) = \sum_{n=1}^\infty \frac{f^{(n)}(0)}{n!} x^n$ (and non-zero radius of convergence) satisfying what you want. That is,
$$
F(x) = \sum_{n=1}^\infty \frac{x^n}{n\cdot n!}.
$$
Deriving this, you get
$$
F'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n!} = \frac{1}{x}\sum_{n=1}^\infty \frac{x^{n}}{n!} = \frac{e^x-1}{x}.
$$
There infinitely many of them... obviously as long as the function is $n$-times differentiable and $f^{(n)}(0)\neq 0$ you can always multiply constant to modify it to be $1/n$. One example is $e^x/n$