While the usual sigmoid function $\sigma(x) = \frac{1}{1+e^{-x}}$ is symmetric around the origin, I'm curious as to whether this generalization of the sigmoid is point symmetric around $(\theta, 0.5)$:
$S(x)=\frac{1}{1+e^{-k(x-\theta)}}$
where the constants $k, \theta > 0$. Is this function indeed point symmetric, and if so, how do you demonstrate it? Is it sufficient to show that $1-S(x)=S(-x)$ since the output is constrained to $[0,1]$?
I also want to be clear what I mean by "point symmetry." Looking at this function geometrically, the output of this function for $x>\theta$ looks like a $180^{\circ}$ rotation around the point $(\theta, 0.5)$ of the function for $x\leq\theta$:
In this graph, $\theta = 50$ and $k = 0.2$. The red line of the first panel is rotated $180^{\circ}$ around the black point $(\theta, 0.5)$ so that it overlays the blue line in the second panel (it has also been shifted up slightly to show the overlap).
First of all, $\sigma$ (as you defined it in your question is point symmetric with respect to $(0,1/2)$ not the origin. This can indeed be shown by $1-\sigma(x)=\sigma(-x)$ which we can rewrite to $$\sigma(x)-\frac{1}{2}=-\left(\sigma(-x)-\frac{1}{2}\right).$$ Hence, the function $\tau(x):=\sigma(x)-\frac{1}{2}$ is indeed point symmetric around the origin.
Now, fix $k$ and let $S(x)=\frac{1}{1+e^{-kx}}$. Then, $$S(x)-\frac{1}{2}=\sigma(kx)-\frac{1}{2}=-\left(\sigma(-kx)-\frac{1}{2}\right)=-\left(S(-x)-\frac{1}{2}\right).$$
Note that your function $S$ (with an additional $\theta$) just translates the function along the $x$-axis, so it is point symmetric around $(\theta,1/2)$ by definition.