Let $(\Omega, \mathcal{F}, \mathbb{P}) = ([-1,1], \mathcal{B}([-1,1]), \frac{1}{2}\lambda|_{\mathcal{B}([-1,1])})$ and let $X: \Omega \to \mathbb{R}$ be given by $X(\omega) = \omega^2$. For $E \in \mathcal{B}([-1,1])$, define $-E := \{x \in [-1,1]: -x \in E\}$.
I want to show that $\sigma(X) = \{ E \in \mathcal{B}([-1,1]): E=-E \}$.
A typical interpretation of $\sigma(X)$ would be the smallest $\sigma$-algebra such that $X$ is measurable. It is straightforward to prove that: \begin{align} \sigma(X) = X^{-1}\mathcal{B}([-1,1]) = \{\{X \in B\} \subset \Omega: B \in \mathcal{B}([-1,1])\}. \end{align}
To prove the claim, the idea should be something like this. Let $\mathcal{G}$ be the $\sigma$-algebra generated by sets of the form $\{ E \in \mathcal{B}([-1,1]): E=-E \}$. Now we must show (1) that $X$ is $\mathcal{G}$-measurable and (2) that $\mathcal{G} \subset \sigma(X)$.
However I do not know if this approach is correct since $X(\omega) = \omega^2$.
It's clear that $\sigma(X)\subseteq \mathcal{C}$, where $\mathcal{C}=\{ E \in \mathcal{B}([-1,1]): E=-E \}$ because $\sigma(X)$ is generated by the sets of the form $\{X\le x\}\in \mathcal{C}$. On the other hand, for any $C\in \mathcal{C}$, $X(C)=X(C\cap [0,1])\equiv A$ is a Borel set ($\because$ the restriction of $X$ to $[0,1]$ is injective) and $\sigma(X)\ni X^{-1}(A)=C$.