Let the eccentricity of hyperbola and conjugate hyperbola be $e_1 \& e_2$respectively. Then,

$\frac{1}{e_1^2}+\frac{{\displaystyle 1}}{{\displaystyle e_2^2}}=1$

$\Rightarrow \frac{1}{{\left({\displaystyle \frac{5}{3}}\right)}^2}+\frac{1}{e_2^2}=1$

$\Rightarrow \frac{9}{25}+\frac{1}{e_2^2}=1$

$\Rightarrow \frac{1}{e_2^2}=1-\frac{9}{25}$

$\Rightarrow \frac{1}{e_2^2}=\frac{16}{25}$

$\Rightarrow e_2=\pm \frac{5}{4}$

But, eccentricity of hyperbola is never less than one. Hence, $e_2=\frac{5}{4}$.