Suppose $X\sim N(0,1)$. Is there a nonlinear smooth function $f:\mathbb R\to\mathbb R$ such that $f(X) + X \sim N(0,1)$?
Here is my attempt. I consider the moment generating function of $X+f(X)$: \begin{equation} \begin{split} \mathbb{E} e^{t(X+f(X))} = \int_{\mathbb{R}} \frac{1}{\sqrt{2\pi}} e^{t x + tf(x)} e^{-x^2/2} dx = e^{t^2/2}, \end{split} \end{equation} which can be further simplified as $$ \int_{\mathbb R} \frac{1}{\sqrt{2\pi}} e^{tf(x)} e^{-(x-t)^2/2} dx = 1. $$ This must hold for any $t\in\mathbb R$. But I don't know how to deal with this integral equation.
Is there any suggestion on this equation or an alternative approach? Thank you!
Update: @Cretin2 Thanks for the suggestion. The density of $Y=f(X) + X$ is $$p_Y(y) = \sum_{x:f(x)+x=y} \frac{p_X(x)}{|f'(x) + 1|},$$ which is reduced to $$ \sum_{x:f(x)+x=y} \frac{e^{f(x)(y-f(x)/2)}}{|f'(x) + 1|} = 1,$$ for any $y\in\mathbb R$. If $f(x)+x$ is not monotone, then there is some point $x_0 < x_1$ such that $f'(x_0) + 1= 0$ and $h:(x_0,x_1)\to\mathbb R, x\mapsto f(x)+x$ is injective. As $y\to h(x_0)$, $$\frac{e^{f(h^{-1}(y)) (y - f( h^{-1}(y) )/2) }}{|f'(h^{-1}(y)) + 1|} \to \infty,$$ a contradiction. Then $f(x)+x$ must be monotone and $f$ must be linear. There is no such nonlinear transformation.
I find https://en.wikibooks.org/wiki/Probability/Transformation_of_Probability_Densities useful (for density after non-injective transformation).