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 $f \in C^1(\mathbb{R})$ with $E[f(Z)]=0$ we have $E[f(Z)^2] \le E[f'(Z)^2]$.

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 $E[f(Z)^2], E[f'(Z)^2] < \infty$. Then there exists a constant $c$ such that $\phi(x) := f(x) + c$ satisfies $E[\phi(Z)^2] = E[\phi'(Z)^2]$.

Proof. Suppose without loss of generality that $E[f(Z)] = 0$. Set $c^2 = E[f'(Z)^2] - E[f(Z)^2]$, which by the Poincaré inequality is nonnegative. Then $$E[\phi(Z)^2] = E[f(Z)^2] + c^2 = E[f'(Z)^2] = E[\phi'(Z)^2].$$

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 Poincaré 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 Poincaré inequality saturates. Again, this can be seen via Hermite polynomials.

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

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

Theorem. For every $f \in C^1(\mathbb{R})$ with $E[f(Z)]=0$ we have $E[f(Z)^2] \le E[f'(Z)^2]$.

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 $E[f(Z)^2], E[f'(Z)^2] < \infty$. Then there exists a unique constant $c$ such that $\phi(x) := f(x) \pm c$ satisfies $E[\phi(Z)^2] = E[\phi'(Z)^2]$.

Proof. Suppose without loss of generality that $E[f(Z)] = 0$. Set $c^2 = E[f'(Z)^2] - E[f(Z)^2]$, which by the Poincaré inequality is nonnegative. Then $$E[\phi(Z)^2] = E[f(Z)^2] + c^2 = E[f'(Z)^2] = E[\phi'(Z)^2].$$

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 $f \in C^1(\mathbb{R})$ with $E[f(Z)]=0$ we have $E[f(Z)^2] \le E[f'(Z)^2]$.

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 $E[f(Z)^2], E[f'(Z)^2] < \infty$. Then there exists a constant $c$ such that $\phi(x) := f(x) + c$ satisfies $E[\phi(Z)^2] = E[\phi'(Z)^2]$.

Proof. Suppose without loss of generality that $E[f(Z)] = 0$. Set $c^2 = E[f'(Z)^2] - E[f(Z)^2]$, which by the Poincaré inequality is nonnegative. Then $$E[\phi(Z)^2] = E[f(Z)^2] + c^2 = E[f'(Z)^2] = E[\phi'(Z)^2].$$