The expression ${\left(\frac{243}{32}\right)}^{-\frac{4}{5}}$ is equal to:

First, express the base terms as powers of base prime numbers:
$$243=3^5$$

$$32=2^5$$

So, we can rewrite the expression:

$$\left(\frac{243}{32}\right)={\left(\frac{3}{2}\right)}^5$$

Now apply the exponent factor of $-\frac{4}{5}$:

$${\left[{\left(\frac{3}{2}\right)}^5\right]}^{-\frac{4}{5}}={\left(\frac{3}{2}\right)}^{5\times \left(-\frac{4}{5}\right)}={\left(\frac{3}{2}\right)}^{-4}$$

Invert the base to turn the negative exponent positive:

$${\left(\frac{2}{3}\right)}^4=\frac{2^4}{3^4}=\frac{16}{81}$$