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I want to show that if $\gamma$ is an isometry of $\mathbb{R}^2$ that preserves orientation, then it is either a translation or a rotation about a point.

If $\gamma$ is an isometry of $\mathbb{R}^2$ then it is of the form $\vec{\gamma}(\vec{x})=A\vec{x}+\vec{b}$ for some orthogonal matrix $A$ and $\vec{b} \in \mathbb{R}^2$. Since $\gamma$ preserves orientation we have that $\det A=1$.

If $A=I$, then clearly $\gamma$ is a translation $\vec{x} \mapsto \vec{x} + \vec{b}$.

I'm stuck with the case $A \neq I$, though. How do I proceed?

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Orthogonal matrices with determinant 1 are exactly rotation matrices (about the origin). Hence $\vec{\gamma}(\vec{x})=A\vec{x}+\vec{b}$ is a rotation plus a translation, i.e. a rotation about the point which has the coordinates of the vector $\vec{b}$.

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    It's rather a rotation that takes $0$ into $b$. – 2017-02-14
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    How would one justify that a rotation plus a translation is indeed a rotation about the point $\vec{b}$? – 2017-02-15
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First observe that translations are rigid motoins or isometry and any rigid motion which fixes the origin preserves the inner product i.e. any rigid motion which fixes the origin is an orthogonal transformation. Now suppose $m$ is an isometry. Then $m$ sends the origin $O$ to somewhere let's say $b$ i.e. $m(O) = b,$ where $b \in \Bbb R^n$. Let $t_{-b}$ be the translation by the vector $-b$ i.e. $t_{-b} (v) = v-b,\ v \in \Bbb R^n$. Let $A=t_{-b} \circ m$. Then clearly $A$ is an isometry since so are $t_{-b}$ and $v$. Also $A(O) = O$ i.e. $A$ fixes the origin. So by our previous observation it follows that $A$ is an orthogonal transformation. Since translations are invertible and $t_{-b} = (t_b)^{-1}$ for all $b \in \Bbb R^n$. So we have $m = t_b \circ A$. So $m(v) = Av + b,\ \text {for all}\ v \in \Bbb R^n$ where $A \in O(n)$.