In fact, every reasonable function can be made into an example by adding an appropriate constant.

I'll write $Z$ for a standard Gaussian random variable.  Recall the Gaussian Poincaré inequality:

> **Theorem.** For every $f \in C^1(\mathbb{R})$ we have $\operatorname{Var}[f(Z)] \le E[f'(Z)^2]$.

Equivalently, this is the fact that the Ornstein-Uhlenbeck "number" operator has spectral gap equal to 1.  Perhaps the simplest way to prove the Poincaré inequality is via Hermite polynomials; see Bogachev, *Gaussian Measures*, Theorem 1.6.4.  The statement generalizes directly to absolutely continuous functions (in the appropriate Sobolev space over Gaussian measure)






> **Corollary.** Let $f \in C^1(\mathbb{R})$ with $E[f(Z)^2], E[f'(Z)^2] < \infty$.  There exist either one or two real numbers $c$ such that $\phi(x) := f(x) + c$ satisfies $E[\phi(Z)^2] = E[\phi'(Z)^2]$.

*Proof.* Set 
$$\begin{align*}\psi(c) &:= E[\phi(Z)^2] - E[\phi'(Z)^2] \\ &= \operatorname{Var}[\phi(Z)] + (E[f(Z)] + c)^2 - E[\phi'(Z)^2] \\
&= \operatorname{Var}[f(Z)] + (E[f(Z)] + c)^2 - E[f'(Z)^2]\end{align*}.$$
Now $\psi(c)$ is a quadratic in $c$ with $\psi(c) \to +\infty$ as $c \to \pm \infty$, and by the Poincar&eacute; inequality we have $\psi(-E[f(Z)]) \le 0$.  So $\psi$ has either one or two real roots.  

Indeed, the only way for the constant $c$ to be unique is if $f$ is a linear function, because that is the only case in which the Poincar&eacute; inequality saturates.  Again, this can be seen via Hermite polynomials.