Let $\alpha$ be a plane curve, whose curvature can be determined using the following formula:
$k_{\alpha}(t) = \frac{det(\alpha^{''}(t),\alpha^{''}(t))}{|\alpha^{'}(t)|^3}$
Let $L$ be a linear map and consider $\beta = L \circ \alpha$. It can be shown easily that $\beta^{n} = L \circ \alpha^{n}$. Thus, the curvature of $\beta$ can be expressed as follows:
$k_{\beta}(t) = \frac{det(\beta^{''}(t),\beta^{''}(t))}{|\beta^{'}(t)|^3} = \frac{det(L(\alpha^{'}(t)),L(\alpha^{''}(t)))}{|L(\alpha^{'}(t)|^3} = \frac{det(L)det(\alpha^{'}(t),\alpha^{''}(t))}{|L(\alpha^{'}(t)|^3} = det(L)k_{\alpha}(t)\frac{|\alpha^{'}(t)|^3}{|L(\alpha^{'}(t)|^3}$
Following @flinty advice, $k_{\beta}(t)$ can be expressed in terms of $k_{\alpha}(t)$. Is there any way to further simplify the expression obtained?