= Solution
Increasing the conserved nucleus density to $10N$ multiplies the low-temperature neutral-gas slope and high-temperature ionized-gas slope by ten. However, at a fixed intermediate temperature, $x^2/(1-x)=S(T)/(10N)$ is smaller, so a smaller fraction is ionized. The curved transition therefore moves to higher temperature; the new graph is not simply ten times the old graph throughout the crossover.
At a fixed chosen <ionization fraction>, $S(T)$ must increase tenfold. Locally this requires
$$
\Delta T\simeq\frac{\ln10}{3/(2T)+R/(k_BT^2)},
$$
which is of order $10^3\,\mathrm K$ in this region. Directly evaluating the same <Saha equation> at $10^{21}\,\mathrm{m^{-3}}$ moves the maximum pressure slope from about $9250$ to $10560\,\mathrm K$. The two asymptotic slopes still have ratio two, because each hydrogen nucleus still contributes one particle when neutral and two when fully ionized.
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