Let $k$ be a local non-archimedean field with ring of integers $\cal O$ and maximal ideal $\frak p$.
Let $\pi$ be an irreducible admissible $\infty$-dimensional representation of $\text{GL}_2(k)$ with central character $\varepsilon$. We know from a classical paper of Casselman that there exists a largest ideal ${\frak p}^s\subset\cal O$ such that the subspace of vectors $v\in\pi$ such that $$ (*)\qquad\qquad\pi(\gamma)v=\varepsilon(a)v\qquad \text{for all }\gamma=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)\in K_0({\frak p}^s) $$ is non-empty and in fact one-dimensional (newvector).
Here $K_0({\frak p}^n)=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)\in{\rm GL}_2(\cal O)$ such that $c\equiv0\bmod{\frak p}^n$.
Incidentally, if (*) occurs then $\varepsilon$ needs to be a character of $K_0({\frak p}^s)$ and this happens only if $s$ is greater then the conductor of $\varepsilon$.
Now consider the contragredient representation $\check\pi$. We know that $\check\pi\simeq\pi\otimes\varepsilon^{-1}(\det)$. It is very easy to check that the central character of $\check\pi$ is $\varepsilon^{-1}$. Indeed $$ \check\pi(zI_2)=\pi(zI_2)\varepsilon^{-1}(\det(zI_2))= \varepsilon(z)\varepsilon^{-1}(z^2)=\varepsilon^{-1}(z). $$ So one would expect, by Casselman's theory, a newvector in $\check\pi$ with character $\varepsilon^{-1}$. But, if $v\in\pi$ is the newvector $$ \check\pi(\gamma)v=\pi(\gamma)\varepsilon^{-1}(det(\gamma))v= \varepsilon(a)\varepsilon^{-1}(ad)v=\epsilon^{-1}(d)v $$ and $\epsilon^{-1}(a)\neq\epsilon^{-1}(d)$. Note that $\varepsilon^{-1}(det(\gamma))=\varepsilon^{-1}(ad-bc)=\varepsilon^{-1}(ad)$ because $ad\in\cal O^\times$ and $\varepsilon$ is trivial on $1+{\frak p}^s$.
So,what is going on?