$\newcommand\colvec[2]{\begin{bmatrix}#1 \\\\ #2\end{bmatrix}} \newcommand\rowvec[2]{\begin{bmatrix}#1\end{bmatrix}} \newcommand\bra[2][]{#1\langle {#2} #1\rvert} \newcommand\ket[2][]{#1\lvert {#2} #1\rangle}$ Suppose $\mathbf{a}\in\mathbb{C}^{2}$ is represented in the standard basis as $\mathbf{a}=\colvec{1}{1}$. In my understanding its dual can be represented as the row vector $\mathbf{\alpha}=\rowvec{1}{1}$, where the components are expressed in the dual basis derived from the standard basis in $\mathbb{C}^{2}$, i.e. $\epsilon_{1}=\rowvec{1}{0}$ and $\epsilon_{2}=\rowvec{0}{1}$. Using the bra-ket notation this seems correct to me: if $\ket{a}=\colvec{1}{1}$, then $\bra{a}=\ket{a}^{\dagger}=\rowvec{1}{1}$.
However, vector components transform in a contravarient fashion under change of coordinates, so if we make the tranformation $e_{1}\rightarrow2e_{1},e_{2}\rightarrow e_{2}$, then $\ket{a}=\colvec{\frac{1}{2}}{1}$. Similarly, the components of dual vectors change in a covarient fashion, so then $\bra{a}=\rowvec{2}{1}$, which isn't $\ket{a}^{\dagger}$ anymore.
Is $\bra{x}$ not the dual of $\ket{a}$ and I'm just confusing naming conventions here, or is there something else I'm missing?