The following question is related to the question:
Let $V$ and $W$ be two vector spaces with scalar products and $\phi: V \to W$ be a linear map with an adjoint $\phi^*$. It is clear, that if $\phi$ and $\phi^*$ are both invertible, then $(\phi^{-1})^*=(\phi^*)^{-1}$. My question is: Does someone have an easy counter-example (to show my students) that the following implication is wrong
$\phi$ is an isomorphism $\Rightarrow$ $\phi^*$ is an isomorphism
e.g. does someone have an isomorphism with adjoint whose inverse doesn't have an adjoint?
The example I have found is the following: Let $V:=\mathbb R^\mathbb N_{fin}$ be the vector space of finite sequences with the inner product
$$ (\mathbf{a}, \mathbf{b}):=\sum_i a_i b_i$$
Then the isomorphism
$$ \phi: V \to V, (a_1, a_2, \dots) \mapsto (a_1, a_2+a_1, a_3+a_2, a_4+a_3, \dots, a_{i}+a_{i-1}, \dots)$$
is invertible and has an adjoint, namely we have:
$$\begin{align*} \phi^{-1}&: (a_1, a_2, \dots) \mapsto (a_1, a_2 -a_1, a_3-a_2+a_1, \dots, \sum_{j=1}^i (-1)^{i-j} a_j, \dots)\\ \phi^{*}&: (a_1, a_2, \dots) \mapsto (a_1+a_2, a_2 +a_3, \dots, a_i+a_{i+1}, \dots) \end{align*} $$ but $\phi^{-1}$ does not have an adjoint, since we would have $$(\phi^{-1})^*(1,0,0, \dots)=(1,-1,1,-1,\dots, (-1)^i, \dots) \not \in V$$ or similarly $\mathbf{e_1}=(1,0,0,\dots)$ does not have a preimage under $\phi^*$.