$E[ f(Z) \cdot Z] =E[Z^2]$ iff $f(z)=z$

Question

Suppose $f(\cdot)$ is a fucntion that is odd, $f(0)=0$, increasing and continuous.Is the following result true

Let $Z \sim \mathcal{N}(0,1)$ then \begin{align} \mathbb E[ f(Z) \cdot Z] =\mathbb E[Z^2] \end{align} iff $f(z)=z$ a.s.

Clearly, if $f(z)=z$ than the statement follows, but is the reverse direction true?

Answer 1

How about $f(z) = \frac{1}{3}z^3$? This gives

$$ \Bbb{E}[f(Z)Z] = \frac{1}{3}\Bbb{E}[Z^4] = 1 = \Bbb{E}[Z^2]. $$

In general, if $g$ is an odd measurable function such that $\Bbb{E}[Z g(Z)]$ is finite and positive ($Zg(Z)$ is always non-negative, so the expectation is well-defined and takes values in $[0, \infty]$), then

$$f(z) = \frac{g(z)}{\Bbb{E}[Zg(Z)]}$$

always satisfies $\Bbb{E}[Zf(Z)] = \Bbb{E}[Z^2]$.