The right one looks like a reflection of the left one along the diagonal line \( y = x \)

If this is true, then they are the inverse function of each other.

So for the right one, we have

\( x = \frac {1} {1 + e^{-\beta y}} \)

\( \Rightarrow 1 + e^{-\beta y} = \frac {1} {x} \)

\( \Rightarrow y = -\frac {1} {\beta} \ln\left(\frac {1} {x} - 1\right) \)

But you may have some further shifts on the curve depends on the location.