Is the dual bundle of canonical line bundle on $\mathbb{CP}^n$ isomorphic to itself?
The canonical line bundle is represented by $\{g_{ij}\}=\{z_jz_i^{-1}\}$, where $g_{ij}$ is the transformation from $U_j$ to $U_i$ on $U_i\cap U_j$.
How to understand the dual bundle $\{g'_{ij}\}=\{g^{-1}_{ij}\}=\{z_iz_j^{-1}\}$?
Is it isomorphic to the canonical bundle?
The bundle you are describing is called $\mathcal O(1)$ (and not the "canonical bundle", which is $\mathcal O(-n-1)$).
Its dual bundle $\mathcal O(-1)$ is called the tautological line bundle .
These bundles are not algebraically nor holomorphically isomorphic because in both categories $$\Gamma(
\mathbb{CP}^n ,\mathcal O(1))=(\mathbb C^{n+1})^*\neq 0= \Gamma( \mathbb{CP}^n ,\mathcal O(-1)).$$
They are not even topologically isomorphic because they have different Chern classes: $$ c_1(\mathcal O(1))=1\neq c_1(\mathcal O(-1)) =-1\in H^2(\mathbb{CP}^n,\mathbb Z)=\mathbb Z$$
This is false already for $n=1$; the canonical bundle of the Riemann sphere has degree $-2$, so its dual has degree $2$. Therefore, they are not isomorphic.
Edit: as Georges points out, I was answering the question in the title, and not the one in the text. I hope that's what you wanted.
