For any random variable (regardless continuous or discrete) that is symmetric about $0$, since $\operatorname{Cov}(X, \operatorname{sign}(X)X^2) = E[|X|^3] \geq 0$, it can be seen that the correlation of your interest is bounded below by $0$. And since $E[|X|^3] = 0$ if and only if $|X|^3 = 0$ almost surely, that is $X = 0$ almost surely, no non-degenerate continuous random variable can attain this $0$ lower bound.
Your second conjecture can be negated by using whuber's comment. Let $X \sim t_\nu$, where $\nu > 4$, it then follows by Theorem 2.2 of MOMENTS OF STUDENT’S T-DISTRIBUTION: A UNIFIED APPROACH by
J. LARS KIRKBY, DANG H. NGUYEN, AND DUY NGUYEN that
\begin{align*}
& \operatorname{Cov}(X, \operatorname{sign}(X)X^2) = E[|X|^3] =
\frac{\nu^{3/2}\Gamma(\frac{\nu - 3}{2})}{\sqrt{\pi}\Gamma(\frac{\nu}{2})}, \\
& \operatorname{Var}(X) = E[X^2] = \frac{\Gamma(\frac{3}{2})}{\sqrt{\pi}}\frac{\nu}{\frac{\nu}{2} - 1} = \frac{\nu}{\nu - 2}, \\
& \operatorname{Var}(\operatorname{sign}(X)X^2) = E[X^4] = \frac{\Gamma(\frac{5}{2})}{\sqrt{\pi}}\frac{\nu^2}{(\frac{\nu}{2} - 1)(
\frac{\nu}{2} - 2)} = \frac{3\nu^2}{(\nu - 2)(\nu - 4)}.
\end{align*}
This gives
\begin{align*}
\operatorname{Corr}(X, \operatorname{sign}(X)X^2) =
\frac{\frac{\nu^{3/2}\Gamma(\frac{\nu - 3}{2})}{\sqrt{\pi}\Gamma(\frac{\nu}{2})}}{\sqrt{\frac{\nu}{\nu - 2}\frac{3\nu^2}{(\nu - 2)(\nu - 4)}}}
= \frac{\Gamma(\frac{\nu - 3}{2})(\nu - 2)(\nu - 4)^{1/2}}{\sqrt{3\pi}\Gamma(\frac{\nu}{2})},
\end{align*}
which converges to $0$ as $\nu \to 4^+$. This implies that there exists a distribution that gives the correlation of your interest arbitrarily close to $0$, implying that a fixed positive lower bound of this correlation does not exist.
