Let the square lamina have side length $a$ and mass $M$.

We compare moments of inertia about different axes lying in the plane of the lamina.

• $AC$ is a diagonal passing through the center $O$.
• $EF$ is the line through the center parallel to side $AD$ (as in the standard square-lamina figure).
• $AD$ is a side axis.

We use symmetry and the perpendicular axis theorem.

1. Moment of inertia about axes through the center in the plane

For a square lamina, the moment of inertia about the axis through the center and parallel to a side is

$I_{EF}=I_x=\frac{Ma^2}{12}$

Similarly, about the perpendicular in-plane central axis,

$I_y=\frac{Ma^2}{12}$

Since the square is symmetric under rotation by $90^\circ$, any axis through the center in the plane has the same moment of inertia.

In particular, the diagonal $AC$ also passes through the center, so

$I_{AC}=\frac{Ma^2}{12}$

Hence,

$I_{AC}=I_{EF}$

So option D is correct.

2. Check the other options

Option A: $I_{AC}=\sqrt{2}\,I_{EF}$

But we found

$I_{AC}=I_{EF}$

So this is false.

Option B: $\sqrt{2}I_{AC}=I_{EF}$

Again, since $I_{AC}=I_{EF}$, this is false.

Option C: $I_{AD}=3I_{EF}$

Using parallel axis theorem, axis $AD$ is parallel to the central axis $EF$ and at distance $a/2$.

So,

$I_{AD}=I_{EF}+M\left(\frac a2\right)^2$

$I_{AD}=\frac{Ma^2}{12}+\frac{Ma^2}{4}$

$I_{AD}=\frac{Ma^2}{12}+\frac{3Ma^2}{12}=\frac{4Ma^2}{12}=\frac{Ma^2}{3}$

Now,

$3I_{EF}=3\cdot \frac{Ma^2}{12}=\frac{Ma^2}{4}$

But

$I_{AD}=\frac{Ma^2}{3}\neq \frac{Ma^2}{4}$

So option C is false.

3. Final answer

The only correct option is

$\boxed{D}$

since

$\boxed{I_{AC}=I_{EF}}$