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Let $X$ be a random variable with $\mathbb{E} X^2 = 1$. Let $X_i$ be i.i.d. copies of $X$ such that $$ \frac{1}{\sqrt{\sum X_i^2}} \left(X_1, ..., X_N\right) $$ is uniformly distributed on $\mathbb{S}^{N-1}$. Prove that $X = \mathcal{N}(0,1)$ in distribution.

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    Does your question mean that $X =N(0,1)$ or as $N$ tends towards infinity $X$ converges in distribution to a $N(0,1)$ random variable. – 2017-02-13
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    It means $X = \mathbb{N}(0,1)$ for fixed $N$. – 2017-02-13
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    A similar question. http://math.stackexchange.com/questions/397116/uniform-distribution-on-the-n-sphere – 2017-02-13
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    That's the converse of what I'm asking. – 2017-02-14

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This isn't super rigorous, but it's an old argument that is historically important (for instance, it was used in deriving the Maxwell-Boltzman distribution, I think.)

We know that the distribution of the vector $\vec X$ must be spherically symmetric, so $$ f_{\vec X}(x_1,\ldots,x_N) = g(x_1^2+\ldots +x_N^2).$$

We also know that the components of the vector are independent and that their distributions are symmetric, so that $$f_{\vec X}(x_1,\ldots,x_N) = h(x_1^2)h(x_2^2)\ldots h(x_N^2)$$ so we have a functional equation $$ g(x_1^2+\ldots +x_N^2) = h(x_1^2)\ldots h(x_N^2).$$

Letting $x_2=0,x_3=0\ldots x_N=0,$ we get $g(x_1^2) = h(x_1^2)h(0)^{N-1}$ so we can write $$ h(0)^{N-1}h(x_1^2+\ldots +x_N^2) = h(x_1^2)\ldots h(x_N^2)$$ or, letting $\tilde h(x) = h(x)/h(0),$ $$ \tilde h(x_1^2+\ldots+x_N^2) = \tilde h(x_1^2)\ldots \tilde h(x_N^2).$$

It is known known that the only nice solution to the functional equation $f(x_1+\ldots +x_N) = f(x_1)\ldots f(x_N) $ is $e^{ax}$ so we have $$ h(x^2) = h(0)e^{-bx^2}$$ and clearly $b$ must be positive for the distribution to be valid.

So finally, we have $$ f_{\vec X}(x_1,\ldots x_n) = C e^{-b(x_1^2+\ldots x_N^2)}$$

So we have $X_i$ is an $N(0,\sigma^2)$ for some variance $\sigma^2.$ Since you have $E(X^2) = 1$ it must be $N(0,1).$