Consider a density function $f(x)$, in one dimension for simplicity. Assume that it has an extremum at $x^*$, so that $f'(x^*) = 0$.
If I make a change of variables via $x = x(\xi)$, then I must consider the transformed density:
$$g(\xi) = f(x(\xi))\frac{\mathrm dx}{\mathrm d\xi}$$
For simplicity I will assume that $x(\xi)$ is a monotonous increasing function.
It is somewhat unintuitive to me that the extremum of $g(\xi)$ is not (in general) at the same spot, i.e., that if $x^* = x(\xi^*)$, then $\xi^*$ is not, in general, an extremum of $g(\xi)$. Mathematically the reason is simple:
$$g'(\xi) = f'(x(\xi))\left(\frac{\mathrm dx}{\mathrm d\xi}\right)^2 + f(x(\xi))\frac{\mathrm d^2x}{\mathrm d\xi^2}$$
Although the first term vanishes at $\xi^*$, the second term need not vanish there.
I find this very unintuitive. It implies that the "density extrema" have no physical meaning, because they depend on the parameterization used to describe the density. However, if $f^*=f(x^*) = 0$, then this location is invariant. That is, extrema that are zero are invariant.
I am looking for examples that make this fact more obvious. Maybe there is a different way to look at it that renders the non-invariance evident? Is there a physical situation where this absence of invariance becomes obvious?
