$I.$ - Assume a cantilever beam and deflection control
$\Delta = P*L^3/3E*I, I = b*t^3/12$, where $\Delta$ is the deflection of a cantilever beam loaded on its free end, P is a unit force, L is the beam length, E is the elastic modulus of the plate, I is the moment of inertia, b is the beam width, and t is the beam depth (plate thickness). Let's plug I into the equation of deflection,
$\Delta = 4P*L^3/b*t^3*E$. In order for the flexibility to remain the same, both beams must be deflected to the same amount, that is
$\Delta_o = \Delta_1$ --->$4P*L^3/b*t_o^3*E_o = 4P*L^3/b*t_1^3*E_1$, the subscription "o" denotes the original plate, and "1" for the new plate. After eliminate the identical parameters of both beams, we get the equation
$t_o^3*E_o = t_1^3*E_1 $, solving for $t_1$ we get
$t_1 = [t_o^3*(E_o/E_1]^{1/3}$
If $E_1 =E_o/2$, $E_o/E_1 = 2$, then
$t_1 = [t_o^3*(2)]^{1/3} = (2)^{1/3}*t_o \approx 1.26*t_o$
$II.$ - Check strength requirement
$M = f_y*S$, in which $f_y$ is the yield strength, and
$S$ is the section modulus, $S = b*t^2/6$,
plug $S$ into the moment equation,
we get $M = f_y*b*t^2/6$.
Let $M_o = M_1$, after eliminating the common constants, we get
$f_{yo}*t_o^2 = f_{y1}*t_1^2$. Solving for $t_1$
$t_1 = t_o^2*\sqrt{(f_{yo}/f_{y1})}$
Check required plate thickness if $f_{y1} = f_{yo}/2$, so
$f_{yo}/f_{y1} = 2$, then
$t_1 = t_o*\sqrt{2} \approx 1.414*t_o$
$III.$ Conclusion
By comparison, $t_1$ derived from the deflection consideration ($1.26*t_o$) is less than that derived from strength concern ($1.414*t_o$), so the latter case governs (if I didn't make mistake in the calculations)./