Is a uniform distribution on a sphere always a norm-scaled normal $mathcal{N}(0,I_d)$ distribution?

Let $U sim Unif(S^{d-1}).$ I was wondering if it’s true that, and if yes, how could we prove that:

$U = frac{Z}{|Z|}$ where $Z sim mathcal{N}(0, I_d), $ i.e. a uniform distribution on a sphere is always a norm scaled distribution? This is kind of like "Polar decomposition in probability."

If yes, one would’ve to just construct $Z.$ To do so, I’d try to use the fact (the "converse statement") that for any $W sim mathcal{N}(0, I_d), frac{W}{|W|}sim Unif(S^{d-1})$ and that ${|W|}sim chi_d,$ a chi distribution with $d$ degrees of freedom. We’d also use that: $|W|$ and $frac{W}{|W|}$ are independent (see Vershynin, Exercise 3.3.7, P.53). More precisely, I’d define $Z:= N U,$ where $Nsim chi_d, U sim Unif(S^{d-1}),$ and enforce the condition that $N, U$ are independent.

If my ideas are so far correct, it remains to show that: 1) $NU sim mathcal{N}(0,I_d)$ and 2) $N =|NU|= |Z|.$ The second one is obvious, since $|U|=1 implies |NU| = N.$ Proving 1) might be a real pain I think, given the complicated PDF of the chi random variable $N$ defined above. So how would we circumvent that problem? Should we some kind of rotational symmetry argument? Of course, if my idea were wrong, then we won’t pursue this route.

P.S. Just a comment: above, we’re trying to construct a normal distribution, given a uniform one. The following might be related but I didn’t find much information on this online, but perhaps a high dimensional version of Box-muller transform is something that’d also transform a uniform distribution into a normal distribution, except in the case, the uniform distribution has to be on an open unit cube, instead of a sphere, unlike the question, which’ll make it simpler, as the co-ordinates would be independent in this case, unlike the question above.