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I'm trying to prove that if $f$ is an orientation preserving isometry of $\mathbb{R}^2$ that doesn't fix any points, then $f$ is a translation.

Suppose $f$ doesn't fix any point. I know that since $f$ is an isometry it must be of the form $Ax+b$. Then since $f$ doesn't fix any point, the equation $Ax-x = (A-I)x=-b$ has no solutions which implies that $\det(A-I)=0$, and therefore $1$ is an eigenvalue of $A$.

If I can deduce that this means $A=I$ then I'm done. Is this true though? And if so, how do I justify it?

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    Look into the other eigenvalue. Which is its absolute value? Can it be negative? – 2017-02-23
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    @amd My apologies, I didn't mean for it to read that way. I meant for them to be completely separate statements. I've made an edit to make it clearer. – 2017-02-23
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    Are you aware of the requirements on the matrix $A$? The matrix $A$ has a particular requirement in order for $Ax+b$ to be an isometry, and also in order for $Ax+b$ to preserve orientation. Without making use of those requirements, I doubt this problem is possible to solve. – 2017-02-23
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    @LeeMosher I'm aware that $A$ has to be orthogonal and that since $f$ is an orientation preserving isometry of $\mathbb{R}^2$ we have $\det{A}=1.$ How do I then use these requirements to solve the problem? – 2017-02-23
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    @Si.0788: Do you know how the determinant and the eigenvalues relate to each other? – 2017-02-23
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    @celtschk Yes, the determinant is the product of the eigenvalues. – 2017-02-23
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    @Si.0788: OK, you have the determinant and one eigenvalue. And you klnow the dimension of $A$. Can you now tell all the eigenvalues? – 2017-02-23
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    @celtschk So the other eigenvalue is also $1$, and therefore $A=I$? – 2017-02-23
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    Yes, except to conclude that $A=I$, you also need to use the fact that $A$ is orthogonal. Otherwise, e.g. $\pmatrix{1&1\\0&1}$ also is a matrix with eigenvalue and determinant $1$. But it's not orthogonal. – 2017-02-23

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In order that $f$ is an isometry, the entries of $A,$ reading from l to r, top to bottom, are either $\cos t, -\sin t, \sin t, \cos t,$ OR $-\cos t, \sin t, -\sin t, -\cos t,$ for some $t\in [0,2\pi).$ In the former case, det($A-I)=0$ implies $\cos t=1,$ and hence $A=I.$ The latter case does not preserve orientation.