16. Given that the perimeter of pentagon $A B C D E$ is 1, and all its interior angles are equal, extend each side outward to intersect at five points forming a star shape. Let $s$ be the perimeter of this star. Then the absolute value of the difference between the maximum and minimum possible values of $s$ is ( ).
(A) 0
(B) $\frac{1}{2}$
(C) $\sqrt{5}$
(D) $\frac{\sqrt{5}+1}{2}$
(E) $\frac{\sqrt{5}-1}{2}$
16. A.

Consider the pentagram as composed of a pentagon and five triangles. Since the interior angles of the pentagon are all equal, each interior angle is
$$
\frac{(5-2) \times 180^{\circ}}{5}=108^{\circ} \text {. }
$$

Therefore, each triangle is an isosceles triangle with base angles of $72^{\circ}$.

Let the sides of the pentagon be $x_{i}(i=1,2,3,4,5)$. Then the perimeter of the pentagram is
$$
\begin{aligned}
s & =2 \times \frac{x_{1}}{2 \cos 72^{\circ}}+2 \times \frac{x_{2}}{2 \cos 72^{\circ}}+\cdots+2 \times \frac{x_{5}}{2 \cos 72^{\circ}} \\
& =\frac{1}{\cos 72^{\circ}}\left(x_{1}+x_{2}+\cdots+x_{5}\right)=\frac{1}{\cos 72^{\circ}} .
\end{aligned}
$$

Thus, regardless of how the pentagon $A B C D E$ changes, as long as the interior angles remain constant, the perimeter $s$ of the pentagram is a constant value $\frac{1}{\cos 72^{\circ}}$.
Therefore, the difference between the maximum and minimum possible values of $s$ is 0.