Let $h: V \to W$ be an isomorphism. How to prove that $h^*$ is also an isomorphism and $(h^*)^{-1}=(h^{-1})^*$?
$h^*$ should be the dual map. Hence, $h^*: W^* \to V^*$ and $h^*(f)=f \circ h$.
I used $h^{-*}=(h^{-1})^*$ and this is the inverse of $h^*$. Thus, $h^* \circ (h^{-1})^*=(h^{-1} \circ h)^*=id^*=id$. But I don't know if this works.
So how to conclude that if $h$ is an isomorphism, the transpose of this linear map $h^*$ is also an isomorphism and $(h^*)^{-1}=(h^{-1})^*$?
It is easy to prove that $(g\circ h)^*=h^*\circ g^*$. Therefore$$(h^{-1})^*\circ h^*=(h\circ h^{-1})^*=\operatorname{Id}^*=\operatorname{Id}$$and$$h^*\circ(h^{-1})^*=(h^{-1}\circ h)^*=\operatorname{Id}^*=\operatorname{Id}.$$So, $(h^*)^{-1}=(h^{-1})^*$.