10. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ always passes through the fixed point $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, and the length of its major axis is in the range $[\sqrt{5}, \sqrt{6}]$. Then the range of the eccentricity is
10. $\left[\frac{\sqrt{3}}{3}, \frac{\sqrt{2}}{2}\right]$.

From the given, $\frac{1}{2 a^{2}}+\frac{1}{2 b^{2}}=1$, which means
$$
a^{2}+b^{2}=2 a^{2} b^{2} \text{. }
$$

Also, $e^{2}=\frac{c^{2}}{a^{2}}=\frac{a^{2}-b^{2}}{a^{2}}$, which means
$$
b^{2}=a^{2}-a^{2} e^{2} \text{. }
$$

From equations (1) and (2), we get
$$
e^{2}=\frac{2 a^{2}-2}{2 a^{2}-1}=1-\frac{1}{2 a^{2}-1} \text{. }
$$

Since $a \in\left[\frac{\sqrt{5}}{2}, \frac{\sqrt{6}}{2}\right]$, therefore, $e \in\left[\frac{\sqrt{3}}{3}, \frac{\sqrt{2}}{2}\right]$.