Step 1: Time period of an oscillating magnet.
A bar magnet swinging in a field $B$ has $T = 2\pi\sqrt{\dfrac{I}{\mu B}}$, where $I$ is its moment of inertia and $\mu$ its magnetic moment. Let a single magnet have period $T_0$.
Step 2: Moment of inertia of the cross.
Two identical magnets stacked and crossed share the same pivot, so their inertias add: $I_{\text{comb}} = I + I = 2I$.
Step 3: Net magnetic moment.
The two moments are perpendicular and equal, so they add as a vector sum: $\mu_{\text{comb}} = \sqrt{\mu^2 + \mu^2} = \sqrt{2}\,\mu$.
Step 4: Period of the combination.
\[ T = 2\pi\sqrt{\frac{2I}{\sqrt{2}\,\mu B}} = 2\pi\sqrt{\frac{\sqrt{2}\,I}{\mu B}} = 2^{1/4}\,T_0, \] since $\dfrac{2}{\sqrt2} = \sqrt2 = 2^{1/2}$ and its square root is $2^{1/4}$.
Step 5: Solve for the single-magnet period.
From $T = 2^{1/4}T_0$, \[ T_0 = 2^{-1/4}\,T. \]
Step 6: State the answer.
Each magnet alone has period $2^{-1/4}T$.
\[ \boxed{2^{-1/4}\,T} \]