$\newcommand{\1}{\mathbf{1}}\newcommand{\E}{\mathbb{E}}\newcommand{\P}{\mathbb{P}}$From the comments I can understand that you have the first part already. One needs to be careful with measurability, though. You have $|X|$ is $\sigma(X^2)$-measurable and hence $\E[|X|\mid\sigma(X^2)]=|X|$ a.s. . We also know $X=|X|$ a.s. by assumption. So it follows (after some straightforward reasonings) that $\E[X\mid \sigma(X^2)]=X$.
Second part. I like your idea for writing $X=|X|\left( \1_{X> 0}-\1_{X<0}\right)$. I use that, but I think there is a shorter way of finishing with that idea, but I'll just go on with mine which a little bit long. As you said in the comments, one has $$\E[X|\sigma(X^2)]=\E[|X|\left( \1_{X>0}-\1_{X<0}\right)\mid \sigma(X^2)]=|X|\E[ \1_{X> 0}-\1_{X<0}\mid\sigma(X^2)]$$ And now we use linearity: $$|X|\E[ \1_{X\geq 0}-\1_{X<0}\mid\sigma(X^2)]=|X|\left(\E[\1_{X> 0}\mid \sigma(X^2)]-\E[\1_{X<0}\mid\sigma(X^2)]\right)$$ We will focus on $\E[\1_{X>0}\mid\sigma(X^2)]$. Just for clarity define
\begin{align}
C:=\{ (X^2)^{-1}(B)\subset \Omega \mid B\in \mathcal B \}
\end{align}
where $\mathcal B$ is the borel $\sigma$-algebra. By definition $\sigma(X^2)=\sigma(C)$. Since $C\subset \sigma(C)=\sigma(X^2)$ we have:
\begin{align}
\int_A\E[\1_{X>0}\mid \sigma(X^2)]\,d\P=\int_A \1_{X>0}\,d\P \ \ \ \ \ \text{ for all } A\in C
\end{align}
For $A\in C$ there is $B\in\mathcal B$ such that $(X^2)^{-1}(B)=A$ so:
\begin{align}
\int_A \1_{X>0}\,d\P&=\P(A\cap X>0)=\P(X^2\in B \ ,\ X>0)
\end{align}
Before going further define for an arbitrary set $U\subset\mathbb R$:
\begin{align}
\sqrt[]{U^+}:=\{ u \in \mathbb R \mid \exists_{v\in U}: v>0 \wedge \sqrt[]{v}=u\}
\end{align}
Now we use symmetry of $X$:
\begin{align}
\int_A 1_{X>0}\,d\P &= \P(X\in \sqrt[]{B^+} \ , \ X>0)\\
&=\P(-X\in \sqrt[]{B^+}\ ,\ -X>0)\\
&=\P(X^2\in B \ , \ X<0)\\
&=\int_A \1_{X<0}\,dP
\end{align}
This holds for all $A$ in $C$.
Notice that $\mu(A):=\int_A \1_{X>0}\,d\P$ and $\nu(A):=\int_A \1_{X<0}\,d\P$ are both measures and equal on a $\pi$-system. It follows by the uniqueness of measure that $\mu(A)=\nu(A)$ for all $A\in \sigma(C)=\sigma(X^2)$. This implies then:
\begin{align}
\int_A\E[\1_{X>0}\mid \sigma(X^2)]\,d\P=\int_A \1_{X>0}\,d\P=\int_A \1_{X<0}\,d\P = \int_A\E[\1_{X<0}\mid \sigma(X^2)]\,d\P
\end{align}
for all $A\in\sigma(X^2)$. By the uniqueness of the conditional expectation we can finally conclude that:
$$ \E[X\mid \sigma(X^2)]=0 \ \ \ \text{ a.s. }$$
