Suppose the rectangle $ABCD$ meets the requirements.

Let $AB=a, AD=b$, and the side of the rhombus obtained after folding is $x$. In the right triangle $ADE$, $x^{2}=b^{2}+(a-x)^{2}$. From this,

$$
x=\frac{a^{2}+b^{2}}{2 a}
$$

By folding the rhombus along its midline and cutting off the parts that are covered only once, we obtained the regular hexagon $EC'D'FA'B'$.

Let's draw the right triangle $ADE$. Since $AD=b$ and $\angle C'EB' = 120^{\circ}$, $\angle DEA = 60^{\circ}$, and $\frac{b}{x} = \sin 60^{\circ}$, we have

$$
x=\frac{2 \sqrt{3}}{3} b
$$

From (1) and (2),

$$
\frac{2 \sqrt{3}}{3} b=\frac{a^{2}+b^{2}}{2 a}
$$

Rearranging the equation, we get $a=\sqrt{3} b$. This condition must be satisfied by the sides of the rectangle for the hexagon to be regular after folding.

It is easy to verify that if the ratio of the sides of the rectangle is such, the hexagon will indeed be regular.
We fold a rectangular piece of paper along its diagonal. After cutting off the parts that exceed the double layer, the remaining paper, when unfolded, will form a rhombus. Now, we fold this rhombus along its midline, and similarly trim the paper as before. What kind of rectangle should we start with so that the final remaining paper, when unfolded, forms a regular hexagon?