Let $b$ be a bilinear form. and consider :
$\psi_L :V\rightarrow V^*$, such that $v \mapsto (f : w \mapsto b(v,w))$
$\psi_R :V\rightarrow V^*$, such that $v \mapsto (g : w \mapsto b(w,v))$
Is it true that $\psi_L$ is an isomorphism iff $\psi_R$ is an isomorphism?
Let $\varphi:V\to V^*$ be a linear map. We define $\varphi': V\to V^*$ by $$ \varphi'(v): w\mapsto \varphi(w)(v).$$ Then you can check that $\psi_R'=\psi_L$ and $\psi_L'=\psi_R$.
Another way of writing $\varphi'$ is $$\varphi': V\to V^{**}\xrightarrow{\varphi^t} V^*$$ where $\varphi^t$ is the transpose of $\varphi$, and the first map is the canonical bidual map, sending $v\in V$ to the evaluation at $v$. Since transposition is functorial, $\varphi^t$ is an isomorphism when $\varphi$ is. So if $\varphi$ is an isomorphism, then $\varphi'$ is an isomorphism iff the bidual map $V\to V^{**}$ is an isomorphism, which happens exactly when $V$ has finite dimension.