Equivalence of inverse transformations under distributional equivalence

Question

Consider continuous, invertible transformations $g,h : \Bbb{R}^d \rightarrow \Bbb{R}^d$ and suppose $g(Y) \overset{d}{=} h(Y)$, where $Y$ is a $N(0, I)$ random variable. Then what can we infer about the transformations $g(\cdot)$ and $h(\cdot)$?

The motivation comes from the one dimensional case, in which we can conclude that either $g^{-1} = h^{-1}$ or $g^{-1} = -h^{-1}$. Essentially, distributional equivalence implies equivalence of the inverse maps. I wanted to know whether something similar exists in the general case. Of course, I don't expect such a "neat" result in higher dimensions due to the spherical symmetry of the standard multivariate gaussian distribution, but anything along these lines would be appreciated.

I am also curious towards the possibility of extending this for a more general class of distributions.

Edit: The question reduces to the following. If $f: \Bbb{R}^d \rightarrow \Bbb{R}^d$ is a continuous, invertible transformation such that $Y \overset{d}{=} f(Y)$, where $Y \sim N(0, I)$, then is it necessary for $f$ to be a rotation, i.e., $f(x) = Rx$ for some orthogonal matrix $R$?

Answer 1

Since the unit standard multivariate normal $\mathcal{N}(0, I)$ is rotationally symmetric, let $R$ be any orthogonal matrix, that is, $R^TR = R R^T =I$, then also $RY \sim \mathcal{N}(0, I)$. Since $g$ and $h$ are invertible $$ g(Y) \stackrel{D}{=} h(Y) \stackrel{D}{=} h(RY) ~\text{using the inverse} \\ Y \stackrel{D}{=} g^{-1}h(RY) $$ so if $g^{-1}h$ is a rotation, we are done, sine product of rotations is a rotation.

This is consistent with what you quote for the one-dimensional case, since the only rotations then are $\pm 1$.