Let $F$ be an orientation-preserving isometry of the euclidean space ($\mathbb{R^2},d$). ( no translation )
Show that: $1)$ $F$ has only one fixed point. So I have to show that there is only one $x \in \mathbb{R^2} $ with $F(x) = x $.
So: First of all $F$ is an orientation-preserving isometry. This means $ F = T_c \circ A $ with $ A \in $ SO(2) (special orthogonal group ) and $T_c$ is a translation. So I want to show $ F(x) = T_c \circ A(x) = Ax + c = x $. $Ax + c$ is equal to ($E_2 - A$ )$x = c$. $E_2$ is the identidy matrix. Moreover $E_2 - A$ is invertible. We get $x= (E_2 - A)^{-1}c$. I'm done, right? Do you see a mistake?
Now Show that : $2)$ there is a $\phi \in (0, 2\pi) $ with $F(y) = x + R_{\phi}(y-x)$ $\forall y \in \mathbb{R^2}.$ $x$ is the fixed point of $F$ , $R_{\phi}$ is the rotation with angle $\phi$. This is the part where I really need your help. I don't know how to solve this. I hope that you help you can help me here.
Your answer to 1) looks good to me.
For 2), take any $y\in\mathbb{R}^2$. Then $F(y)=Ay+c$, and since $x$ is the fixed point, $x=F(x)=Ax+c$, so you need to show $A y+c=Ax+c+R_{\phi}(y-x)$, or $A(y-x)=R_{\phi}(y-x)$. But elements of $SO(2)$ are rotations, so just take $R_{\phi}=A$.