= Solution
Assume the $g$ cycle has least period $n$ and nonzero <periodic-orbit multiplier> $P=(g^n)'(\hat x)$. It then avoids zero. Combining the iterate identity with the <periodic-orbit multiplier>-sign formula gives
$$
f_L^n(\hat x)=\operatorname{sgn}(P)\hat x.
$$
Thus $P>0$ makes $\hat x$ return after $n$ iterates, whereas $P<0$ gives $f_L^n(\hat x)=-\hat x$. In the latter case oddness on the noncritical orbit implies
$$
f_L^{2n}(\hat x)=f_L^n(-\hat x)=-f_L^n(\hat x)=\hat x.
$$
To prove the least periods, note that a $g$ cycle cannot contain two distinct points $x$ and $-x$: evenness would give them the same image, but $g$ permutes its distinct cycle points bijectively. Therefore the $n$ absolute values in the cycle are distinct. Since $|f_L^k(\hat x)|=|g^k(\hat x)|$, any return under $f_L$ must occur at a multiple of $n$. For $P<0$ the first such return fails because $\hat x\ne0$, so $2n$ is minimal. Hence the <period transfer from a quadratic map to its Lorenz sign lift> is
$$
\boxed{P>0:\text{ least period }n;\qquad P<0:\text{ least period }2n.}
$$
There are two symmetry-related $f_L$ cycles in the positive case and one cycle containing both signs in the negative case. Away from zero $f_L'(x)=2|x|$, so their <periodic-orbit multipliers> are $|P|$ and $|P|^2$, respectively. A stable $g$ cycle therefore gives stable lifted cycles in both cases.
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