$rm SU(2)$ transformation of spinors

In the book QFT by Ryder on the topic $rm SU(2)$ and the rotation group, it is stated that,

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A spinor $xi$ transforms under $rm SU(2)$ as,

$$xi rightarrow xi’ = U xi, quad xi^dagger rightarrow xi’^dagger = xi^dagger U^dagger, qquad xi = begin{bmatrix}xi_1 xi_2end{bmatrix}$$

which we see that $xi$ and $xi^dagger$ transform in different ways but we may use the unitarity of $U$ to show that,

$$begin{bmatrix}xi_1 xi_2end{bmatrix}, qquad begin{bmatrix}-xi_2^* xi_1^*end{bmatrix}$$

transform in the same way under $rm SU(2)$. The matrix $U$ is given by,

$$U = begin{bmatrix}a & b -b^* & a^*end{bmatrix}.$$

Applying $U$ on those two spinor representations,

begin{align}
xi’_1 & = a xi_1 + b xi_2
xi’_2 & = -b^* xi_1 + a^* xi_2
end{align}

begin{align}
-xi’^*_2 & = a (-xi^*_2) + b xi^*_1
xi’^*_1 & = -b^* (-xi^*_2) + a^* xi^*_1
end{align}

Now,

$$begin{bmatrix}-xi^*_2 xi_1^*end{bmatrix} = begin{bmatrix}0 & -1 1 & 0end{bmatrix} begin{bmatrix}xi^*_1 xi_2^*end{bmatrix} = zeta xi^*, qquad qquad zeta = begin{bmatrix}0 & -1 1 & 0end{bmatrix}$$

so we have shown that $xi$ and $zeta xi^*$ transform in the same way under $rm SU(2)$.

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First of all, how did he get $begin{bmatrix}-xi_2^* xi_1^*end{bmatrix}$ and knows that it will transform in the same way with $begin{bmatrix}xi_1 xi_2end{bmatrix}$ under SU(2)?

Second, how can applying $U$ on the spinor representations above show that $xi$ and $zeta xi^*$ transform in the same way? I do not see what he is showing there that corresponds to his conclusion.