jmc

Let the original radius and height be $r$ and $h$ respectively, so the original volume is $\pi r^{2}h$.

The new radius and height are $\frac{4}{5}r$ and $\frac{5}{4}h$ respectively, so the new volume is $\pi {\left(\frac{4}{5}r\right)}^2\frac{5}{4}=\frac{4}{5}\pi r^{2}h$, which is $20\%$ less than the original volume.

Hence the desired percent change is $\boxed{20}$ percent.