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I stumbled upon but absolutely no idea how to do It since i'm not keen on group-theory :

Find the rotation group and the isometry group leaving $\gamma(t)=(\dfrac{1}{2}(2cos(t)-cos(2t)),\dfrac{1}{2}(2sin(t)-sin(2t))$ invariant.

I plotted the curve and calculated its curvature but it doesn't seem to be relevant

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    When you plot it can you see any rotational symmetry? Any reflection symmetry? – 2017-02-21
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    By changing $t\mapsto -t$ there's a reflection symmetry along the "x(t)" axis and a rotationnal symmetry by changing $t\mapsto t+2\pi$. So the rotation group is generated by $t+2k\pi, k\in \mathbb{N}$ ? – 2017-02-21

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