6. Given complex numbers $z_{1}$ and $z_{2}$ satisfy $\left|z_{1}\right|=\left|z_{1}+2 z_{2}\right|$, and $\overline{z_{1}} z_{2}=a(2-\mathrm{i})$ (where $a$ is a non-zero real number). Then $\frac{z_{2}}{z_{1}}=$
6. $-\frac{4}{5}+\frac{2}{5}i$.

From the problem, we have
$$
\begin{array}{l}
z_{1} \overline{z_{1}}=\left(z_{1}+2 z_{2}\right)\left(\overline{z_{1}}+2 \overline{z_{2}}\right) \\
\Leftrightarrow 4 z_{2} \overline{z_{2}}+2\left(z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}\right)=0 .
\end{array}
$$

Combining $z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}=4 a$, we get
$$
\left|z_{2}\right|^{2}=-2 a \text {. }
$$

From $\overline{z_{1}} z_{2}=a(2-\mathrm{i}) \Rightarrow\left|z_{1}\right|=\frac{|a(2-\mathrm{i})|}{\left|z_{2}\right|}$
$$
\Rightarrow\left|z_{1}\right|^{2}=-\frac{5}{2} a \text {. }
$$

Thus, $\overline{z_{1}} z_{2}=a(2-\mathrm{i}) \Rightarrow z_{1} \overline{z_{1}} z_{2}=a(2-\mathrm{i}) z_{1}$
$\Rightarrow \frac{z_{2}}{z_{1}}=\frac{a(2-\mathrm{i})}{\left|z_{1}\right|^{2}}=-\frac{2}{5}(2-\mathrm{i})$
$=-\frac{4}{5}+\frac{2}{5} \mathrm{i}$.