Let's take the spin operator along a unitary vector  $\hat{\vec \sigma} \cdot \vec n $. We can find the spin eigenstate with eigenvalue $\hbar/2$. If we rotate the system, $\vec n$ is transformed by a $SO(3)$ matrix and (since the spin eigenstate depends on $\vec n$) the spin eigenstate transforms consequently. If we search the matrices that transforms the spin eigenstate as described above, will we find the matrices of $SU(2)$?
I tried and I didn't find the $SU(2)$ matrix that I expected

The spin operator in the direction of the unitary vector $\vec n$ is $$\hat {\vec \sigma} \cdot \vec n=\hbar/2
\begin{bmatrix}
n_z & n_x-in_y\\
n_x+in_y & -n_z
\end{bmatrix}
$$
Doing some calculation I found that the eigenstate with eigenvalue $\hbar/2$ of this operator is

$$e^{i\phi}\sqrt {\frac {1-n_z}2}\begin{bmatrix} \frac {-n_x+in_y}{n_z-1}\\ 1\end{bmatrix}$$
except for the case $n_z=1$ in that case it is $e^{i\phi} \begin{bmatrix}1\\0\end{bmatrix}$ where $\phi$ can be every real value

Now, if I rotate the unitary vector $\vec n$ of an angle $\theta$ around the z axis it will change in this way $$\begin{bmatrix}n'_x\\n'_y\\n'_z\end{bmatrix}=\begin {bmatrix} cos\theta & sen\theta & 0\\-sen\theta & cos\theta & 0\\0 &0&\ 1 \end {bmatrix} \begin{bmatrix}n_x\\n_y\\n_z\end{bmatrix}$$ thus the spin state will change in this way
$$e^{i\phi}\sqrt {\frac {1-n'_z}2}\begin{bmatrix} \frac {-n'_x+in'_y}{n'_z-1}\\ 1\end{bmatrix}=\begin{bmatrix} cos\theta+isen\theta & 0 \\ 0&1\end{bmatrix}e^{i\phi}\sqrt {\frac {1-n_z}2}\begin{bmatrix} \frac {-n_x+in_y}{n_z-1}\\ 1\end{bmatrix}$$
so the matrix $$\begin{bmatrix} cos\theta+isen\theta & 0 \\ 0&1\end{bmatrix} $$ is the matrix that rotate the spin state when system is rotated. This matrix isn't the one that rotates spinors around the z axis and this confuses me, am I wrong with the calculations or is the idea wrong?