Find the general $f^n(x)$ where
$f^1(x)=e^x$
$f^{a}(f^{b}x)=f^{(a+b)}(x)$
where $n,x\in\mathbb R$
I'm fairly confident that no two $f^n(x)$ with different values of $n$ will intersect, since then we could use the argument that if $f^n(x)=x$ at some point $A$, then $f^m(x)=x$ at the same point $A$, and $x$ does not intersect $e^x$. So $f^n(x)\ne f^m(x)$ at any point.
P.S. I'm not sure what tags to add for this, so add them as you see fit
One may apply a Newton series as this answer shows to yield:
$$f^\alpha(x)=f^{\lceil\alpha\rceil}(f^{\alpha-\lceil\alpha\rceil}(x))=f^{\lceil\alpha\rceil}\left(\sum_{m=0}^{\infty} \binom {\alpha-\lceil\alpha\rceil}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^k(x)\right)$$
other neat forms:
$$f^\alpha(x)=f^{\lceil\alpha\rceil}\left[\lim_{n\to\infty}\binom {\alpha-\lceil\alpha\rceil}n\sum_{k=0}^n\frac{\alpha-\lceil\alpha\rceil-n}{\alpha-\lceil\alpha\rceil-k}\binom nk(-1)^{n-k}f^k(x)\right]$$
note the use of $\alpha-\lceil\alpha\rceil$ to avoid the calculation of half iterations of functions that are super linear with base greater than $e^{1/e}.$
