Are centered Gaussian densities given by $$f_X(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-x^2/(2 \sigma^2)}$$ the unique densities such that the joint pdf of $N > 1$ independent and identically distributed random variables with density $f_X$ is isotropic, that is, rotation invariant?
Thus, one asks that, for some valid function $h$, $$\prod_{i=1}^Nf_{X}(x_i) = h(R),\qquad R = \sqrt{x_1^2+...+x_N^2}$$ This question came up in a discussion and I found it very interesting. I have attempted to solve it as follows.
Note first that $f_X$ must be symmetric, otherwise a rotation mapping $x_i$ to $-x_i$ would violate the symmetry. Therefore, without loss of generality, we may assume that $$f_X(x) = c e^{-g(x^2)}$$ for some function $g$ such that $g(0)=0$ and some normalizing constant $c$. Then we need the joint pdf to satisfy $$f_X(x_1)\cdots f_X(x_N) = C e^{-h(R^2)}$$ for some function $h$ and some constant $C$, such that, still without loss of generality, $h(0)=0$. From this we get $$\sum_{i=1}^N{g(x_i^2)}=h\left(\sum_{i=1}^N{x_i^2}\right)$$ Considering $x_i=0$ for $i \neq 1$, one gets $$g(x_1^2) + (N-1) g(0) = h(x_1^2)$$ $$\therefore g(x) = h(x), \forall x > 0$$ Considering instead $x_i=0$ for $i \notin \{1,2\}$ and assuming $N > 1$, one gets $$g(x_1^2)+g(x_2^2) = h(x_1^2+x_2^2)$$ $$g(x)+g(y) = g(x+y), \forall x,y > 0$$ $$\therefore g(x) = \lambda x, \forall x > 0$$ Thus, substituting $g$ back into $f_X$ implies that the distribution is centered Gaussian: $$f_X(x) = c e^{-\lambda x^2}$$ Now, the observation that centered Gaussian distributions are indeed isotropic completes the proof.
Is the proof sound? I'm looking forward to correcting any mistakes it may have. I also welcome your own proof, or a link to a proof. Thanks in advance.