Chọn phương án D.

$\begin{array}{lll}
&\left|(z-4)\big(\overline{z}-4i\big)\right|&=|z+4i|^2\\
\Leftrightarrow&\left|(z-4)\big(\overline{z}+\overline{4i}\big)\right|&=|z+4i|^2\\
\Leftrightarrow&\left|(z-4)\cdot\overline{(z+4i)}\right|&=|z+4i|^2\\
\Leftrightarrow&|z-4|\cdot|z+4i|-|z+4i|^2&=0\\
\Leftrightarrow&|z+4i|\cdot\big(|z-4|-|z+4i|\big)&=0\\
\Leftrightarrow&\left[\begin{array}{ll}|z+4i|=0 &(1)\\ |z-4|=|z+4i| &(2)\end{array}\right.
\end{array}$

• $(1)\Leftrightarrow z+4i=0\Leftrightarrow z=-4i$.
  Khi đó $\big|z^2\big|=2\big|z-\overline{z}\big|=16$ (đúng).
• Đặt $z=x+yi$, khi đó $$\begin{aligned}
  (2)&\Leftrightarrow|(x-4)+yi|=|x+(y+4)i|\\ &\Leftrightarrow\sqrt{(x-4)^2+y^2}=\sqrt{x^2+(y+4)^2}\\ &\Leftrightarrow(x-4)^2+y^2=x^2+(y+4)^2\\ &\Leftrightarrow-8x=8y\\ &\Leftrightarrow x=-y\,\,(3).\end{aligned}$$
  Ngoài ra, $\begin{aligned}[t]\big|z^2\big|=2\big|z-\overline{z}\big|&\Leftrightarrow\big|x^2-y^2+2xyi\big|=2\big|2yi\big|\\ &\Leftrightarrow\big|-2y^2i\big|=4\big|yi\big|\\ &\Leftrightarrow\big|y^2\big|=2\big|y\big|\\ &\Leftrightarrow\left[\begin{array}{l}y=0\\ y=\pm2\end{array}\right.\end{aligned}$

Vậy có $4$ số phức thỏa đề là $-4i$, $0$, $2-2i$, $2+2i$.