One. (20 points) The two roots of the equation $x^{2}-2 \sqrt{2} x+1=0$ are $\alpha$ and $\beta$. Find the quadratic function $f(x)$ that satisfies $f(\alpha)=$ $\beta, f(\beta)=\alpha, f(1)=1$.
Given $\alpha+\beta=2 \sqrt{2}, \alpha \beta=1(\alpha \neq \beta)$.
Let $f(x)=a x^{2}+b x+c$. From the conditions, we have
$$
\begin{array}{l}
a \alpha^{2}+b \alpha+c=\beta, \\
a \beta^{2}+b \beta+c=\alpha, \\
a+b+c=1 .
\end{array}
$$
Adding (1) and (2) and applying $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=6$

we get
$$
a\left(a^{2}+\beta^{2}\right)+b(\alpha+\beta)+2 c=\alpha+\beta,
$$

which simplifies to $3 a+\sqrt{2} b+c=\sqrt{2}$.
Subtracting (2) from (1) yields
$$
\begin{array}{l}
a\left(\alpha^{2}-\beta^{2}\right)+b(\alpha-\beta)=\beta-\alpha, \\
a(\alpha+\beta)+b=-1, \\
2 \sqrt{2} a+b=-1 .
\end{array}
$$
Solving (3), (4), and (5) together, we get
$$
a=1, b=-(2 \sqrt{2}+1), c=2 \sqrt{2}+1 \text {. }
$$

Thus, $f(x)=x^{2}-(2 \sqrt{2}+1) x+2 \sqrt{2}+1$.