When do $\phi^2$ and $\phi’^2$ have the same expectation under a Gaussian random variable? I am looking for a function $\phi(x)$ such that
$\mathbb{E}_{x\sim\mathcal{N}(0,1)}[\phi(x)^2] = \mathbb{E}_{x\sim\mathcal{N}(0,1)}[\phi'(x)^2]$.
Obvious solutions are $\phi(x) = x$ and $\phi(x) = \exp(x)$. But do you know any other non-trivial solution?
 A: 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.
A: A family of such functions is given by 
$$\phi(x)=\phi_c(x):=e^{t_cx+cx^2}$$
for all real $x$, where $c\in[-\frac{1+\sqrt2}2,\frac{-1+\sqrt2}2]$ and $t_c=\pm\sqrt{1 - 8 c + 12 c^2 + 16 c^3}$; then both expected values equal 
${e^{{2 t_c^2}/(1-4 c)}}/{\sqrt{1-4 c}}={e^{{2 - 8 c - 8 c^2}}}/{\sqrt{1-4 c}}$. 
In particular, choosing here $c=0$, we get two members of this family, given by $\phi_0(x)=e^{\pm x}$ for all real $x$. 
Another two members of this family are given by $\phi_c(x)=e^{cx^2}$ for $c\in\{-\frac{1+\sqrt2}2,\frac{-1+\sqrt2}2\}$ and all real $x$. 
A: Another family of examples, now parametrized by an arbitrary function in a certain general class of functions: 
Let $g\colon\mathbb R\to\mathbb R$ be any bounded continuously differentiable function with a bounded derivative such that $g(0)\ne0$ and $g'(u)^2\ge1$ if $|u|\le1$. Let us then show that a function $\phi$ of the form $f_c$ for some real $c>0$ will do, where 
$$f_c(x):=g(cx)$$
for all real $x$; that is, for some real $c>0$ we will have 
$$Eg(cZ)^2=c^2Eg'(cZ)^2, \tag{1}$$
where $Z\sim N(0,1)$. 
Let 
$$D(c):=Eg(cZ)^2-c^2Eg'(cZ)^2.$$
Then, by dominated convergence, $D(c)$ is continuous in $c\ge0$. Also,
$$D(0)=g(0)^2>0.$$
On the other hand, for $c\to\infty$
$$Eg'(cZ)^2\ge Eg'(cZ)^2 1_{|Z|\le1/c}\ge E1_{|Z|\le1/c}=P(|Z|\le1/c)\gtrsim\frac2{c\sqrt{2\pi}},$$
whereas $Eg(cZ)^2$ stays bounded (since $g$ is bounded). So, $D(c)\to-\infty$ as $c\to\infty$. 
Therefore and because $D(c)$ is continuous in $c\ge0$, we see that $D(c)=0$ for some real $c>0$, which indeed yields (1). 

Clearly, the condition $g'(u)^2\ge1$ if $|u|\le1$ here can be relaxed just to $g'(0)\ne0$. 
E.g., we may take $g(x)=1+\sin x$ for all real $x$, and then 
$D(c)=\frac{3}{2}-\frac{1}{2} e^{-2 c^2}-e^{-c^2} c^2 \cosh \left(c^2\right)$, and the equation $D(c)=0$ has two roots, $c\approx\pm1.73$. 
