I was glad to do it.
But this is not quite what I meant.
I meant: The Dalai Lama is visible in the full size image, disappears in the linear light toolchain image, and remains in the sRGB toolchain image.
I believe that what you seek is impossible, at least by any method resembling what Eric did.
Let me give a brief explanation as to how Eric's original image was constructed:
Given an input color, c, we seek two new colors, x and y, such that
(x + y)/2 = constantGray
((x^gamma + y^gamma)/2)^(1/gamma) = c
We can solve for x and y using numerical methods, unless we simplify using a gamma of 2, in which case there is a closed form solution. In any case, since either solution involves taking roots, we have to be careful about the values allowed to c. Furthermore, we want x and y to be limited to [0, 1]. As it turns out, if we assume that c is in [0, 1], we can solve both problems by applying a linear transformation to c.
Now, if we wish to apply the same thinking to the reverse situation in which the Dalai Lama disappears from the image scaled via linear light and is visible in the image scaled via nonlinear light, we simply swap c and constantGray.
(x + y)/2 = c
((x^gamma + y^gamma)/2)^(1/gamma) = constantGray
The first equation cannot be changed because it ensures that the Dalai Lama will be visible in the nonlinear light downsized image. Furthermore, the second equation cannot be changed because it ensures that the Dalai Lama will disappear from the linear light downsized image. These are the two equations I used in the construction of the images in my previous post (using a gamma of 2.2).
Therefore, if we seek only two colors to replace each original color, the unscaled image will always appear gray. The only possibility I can think of to obtain what you want would be to replace each original color with three (or more) colors, but then I think the unscaled image would start to appear more and more convoluted. Even then, if you were to stand far enough away from your screen, I believe the unscaled image would still very likely appear as gray.
If you have any ideas how one might obtain the results you're looking for, I'd be very interested to read them.