The key part of the question is given sufficient energy. The difference in electronegativity between Ca and Mg is ~0.3 eV. The relationship of eV:K is about 1:12,000. Given that, you should be able to calculate a temperature at which the reaction $\ce{Ca(OH)2 + Mg -> Mg(OH)2 + Ca}$ is favored over the reverse.
However, temperature is based on the random motion of particles, so the energy distribution follows a Gaussian curve. Some Mg atoms are sure to have enough energy to knock loose some Ca atoms at room temperature.
Furthermore, you can force the suggested reaction by making use of relative solubilities. Find a solvent (not water, since magnesium and calcium are far more electronegative than hydrogen, and would decompose water to form hydroxides of both metals) in which $\ce{Ca(OH)2}$ is more soluble than $\ce{Mg(OH)2}$. Precipitation will "lock up" the $\ce{Mg(OH)2}$ as it forms, little by little, forcing the reaction to proceed even though it is endothermic.
Of course, you could also force this to proceed in molten salts by driving it with an electric current.