Let $A$ be an invertible matrix, and let $A = W\Sigma V^*$ be its singular value decomposition. Find a singular value decomposition for $A^*$ and $A^{-1}$.
If we take the conjugate transpose, we get $A^*=V\Sigma^* W^*=V\Sigma W^*$ since $\Sigma$ only has real diagonal entries. But how do we know the singular values of $A$ are the singular values of $A^*$?
Similarly, we have $A^{-1}=V\Sigma^{-1}W^*$. But how do we know the singular values of $A$ are the singular values of $A^{-1}$?
From the definition of singular-value decomposition, singular values are the diagonal entries $\sigma_i$ in $\Sigma$.
You already gave a decomposition $A^*=V\Sigma W^*$. Since $A$ and $A^*$ have the same $\Sigma$ in their decompositions, they have the same singular values.
You also gave a decomposition $A^{-1}=V\Sigma^{-1}W^*$. So if $\sigma_i$ is a singular value of $A$, then $1/\sigma_i$ is the singular value of $A^{-1}$. Don't worry about $\sigma_i=0$. It does not happen because $A$ is non-singular.