How to determine A? I'm struggling with the following problem:
Let $(\gamma, \tau)$ be an arc-length parametrized curve and $\mathcal A: \mathbb R^2 \to \mathbb R^2$ be a Euclidean transformation so that the curve $\hat \gamma := \mathcal A\gamma$ satisfies $\hat \gamma(0)=0$ and $\hat \gamma'(0)= (1,0)$.
My question: How to determine $\mathcal A$ in terms of $\gamma(0)$ and $\gamma'(0)$ now? 
And what exactly is a Euclidean-transformation?
A hint would be much appreciated.
 A: Let $\gamma(0)=:{\bf a}=(a_1,a_2)$ and $\gamma'(0)=(\cos\alpha,\sin\alpha)$ (note that you have assumed $|\gamma'(\tau)|\equiv1$). The euclidean transformation ${\cal A}$ you are after is the composition of the translation $$T_{-{\bf a}}:\quad {\bf x}\mapsto{\bf x}-{\bf a}$$ with the rotation $R_{- \alpha}$ around ${\bf 0}$. The latter is given by the matrix
$$[R_{-\alpha}]=\left[\matrix{\cos\alpha&\sin\alpha\cr-\sin\alpha&\cos\alpha\cr}\right]\ .$$ 
We now have to express ${\cal A}=R_{-\alpha}\circ T_{-{\bf a}}$ in terms of coordinates in order to obtain the explicit parametrization of $$\hat\gamma={\cal A}\circ\gamma=R_{-\alpha}\circ T_{-{\bf a}}\circ\gamma\ .$$ 
Obviously
$$T_{-{\bf a}}\circ\gamma(\tau)=\left[\matrix{\gamma_1(\tau)-a_1\cr \gamma_2(\tau)-a_2\cr}\right]\ ,$$
hence
$$\hat\gamma(\tau)=R_{-\alpha}\circ T_{-{\bf a}}\circ\gamma(\tau)=\left[\matrix{\cos\alpha&\sin\alpha\cr-\sin\alpha&\cos\alpha\cr}\right]\>\left[\matrix{\gamma_1(\tau)-a_1\cr \gamma_2(\tau)-a_2\cr}\right]=\ldots\ .$$
On the RHS we don't obtain a $2\times2$-matrix, but a column vector, as it should be.
