Let $\gamma$ and $\gamma˜$ be two plane curves. Show that, if $\gamma˜$ is obtained from $\gamma$ by applying an isometry M of $\mathbb{R^2}$, the signed curvatures $κs$ and $˜κs$ of $\gamma$ and $\gamma˜$ are equal if M is direct but that $˜κs = −κs$ if M is opposite (in particular, $\gamma$ and $\gamma˜$ have the same curvature). Show, conversely, that if $\gamma$ and $\gamma˜$ have the same nowhere-vanishing curvature, then $\gamma˜$ can be obtained from $\gamma$ by applying an isometry of $\mathbb{R^2}$.
I know that a direct isometry means a function M such tha $M(v)=Pv+a$ where $P$ is an orthogonal Matrix with Det=1. What i thought of is that Since i know $\gamma˜$ is obtained by a such Isometry means $γ˜=M(\gamma)=P\gamma+a$ also i know that the signed curvature is $ks=y''n$ where n is regural to $y'$. so $M(ks)=M(y''n)=P(y''n)+a$ why that is equal to $ks$ i cant get to that. I'm just learning diff.geometry self studying so if it is possible to explain in a more detailed way if it is possible.