Let's consider a smooth positive bounded and non-increasing function $h$ over $\mathbb{R}^{+}$ (for example some kind of decreasing sigmoid).
A) Is it true that if $h$ has only one inflexion point, then so does $h\circ h$? Any proof or counter-example would be much welcome.
B) If $h = f \circ g$ where $f,g$ are positive sigmoid-like (smooth and bounded) functions on $\mathbb{R}^{+}$ with a single inflexion point and $f$ non-decreasing and $g$ non-increasing, then can $h$ or $h\circ h$ have more than one inflexion point?
Take $g(x)=-\frac{7 x^3}{2}+\frac{21 x^2}{2}-\frac{85 x}{8}+\frac{33}{8}$, and set $f(x)=\begin{cases}g^{-1}(x)&\text{if $0\le x\le 4$}\\ h(x)&\text{if $x>4$}\end{cases}$
We choose $h$ without inflexion points, positive and decreasing, such that $f$ is smooth. This is possible, since $g^{-1}(4)\approx 0.0119041928166$ and $(g^{-1})'(4)\approx -0.0963716$. Moreover $g^{-1}$ is decreasing and has only one inflexion point, namely $x=1/2$, $g^{-1}(x)=1$, hence $f$ satisfies the required properties.
But $f\circ f$ has three inflexion points, at $x_1=0.501059728212457, f(f(x_1))=0.5030949274084365$,
$x_2=0.7056987251370224$, $f(f(x_2))=0.6911169347561876$ and
$x_3=0.9990219172667842$, $f(f(x_3))=0.9971533801370549$.
This can be seen, considering that $f([0,4])=[0.0119,1.5]$ and $f(f([0,4]))\sim[0.35943651640802904,1.4956349330112384]$, and so $(f\circ f)(x)$ is given by $(g^{-1}\circ g^{-1})(x)$ for $x$ in that interval.
Now $g\circ g$ is strictly increasing in the interval $[0.35943651640802904,1.4956349330112384]$ and has three inflexion points, hence each of the inflexion points yields an inflexion point of $f\circ f$.
Note that all numerical values are approximate.