If $X$ is a symmetric $n$-dimensional random vector with mean $0$ then is it true that: \begin{align*} & X \text{ follows a multivariate normal law} \\ & \text{iff} \\ & \|X\| \text{is a chi random variable with $n$ degrees of freedom?} \end{align*}
One implication is direct but is the converse true? That is does $\|X\|\sim\chi$ imply $X$ is normal?
Explanation of Symmetric
A random variable is said to be symmetric iff:
For each $i,j\in \{1,\ldots,n\}$ $X_i=X_j$.
In other words: $X=\sum_{i=1}^n X_1e_i$, where $e_i$ are the standard basis vectors of $\mathbb{R}^n$.
Here is the current edition of the question:
Let $X=(X_1,\ldots,X_n)$. As phrased above, the question ignores all questions of the correlations between $X_i$, $X_j$ and the variances.
Suppose, for example, that $X_1,\ldots,X_n$ are independent (which doesn't follow from the assumptions above) and normally distributed with mean $0$ and respective standard deviations $\sigma_1,\ldots,\sigma_n$. Then $$ \sum_{i=1}^n \frac{X_i^2}{\sigma_i^2} \sim \chi^2_n, $$ but the same is not true of $\displaystyle\sum_{i=1}^n X_i^2$.
Next, suppose $Y_1,\ldots,Y_n$ are independent, are normally distributed, have expectation $0$, and have variance $1$, and let $$ X_i = Y_i - \overline Y, \quad \text{where } \overline Y = \frac{Y_1+\cdots+Y_n} n. $$ Then one can show that $(X_1,\ldots,X_n)$ has a multivariate normal distribution, that each $X_i$ has expected value $0$, that they all have the same standard deviation, and even that all pairs $(X_i,X_j)$ (for $i\ne j$) have the same distribution (and are negatively correlated), and that $$ \sum_{i=1}^n X_i^2 \sim \chi^2_{n-1}. $$