Cartan's Signature of curve after Euclidean and Equi-affine transform

Question

Let's start with definitions:

For given curve $C(x(t),y(t))$ in $R^2$ when $t\in[a,b]$.

Cartan's signature that defined in the following way: $\{\kappa.\kappa_s\}$ when $\kappa$ is invariant curvature of the curve, $s$ is invariant acrlength and $\kappa_s$ is derived of curvature by acrlength. Let's denote Cartan's signature of curve $C$ as $S(C)$.

Euclidean/Equi-affine transform is transform from $R^2$ to $R^2$ such $T(v) = A v + t$ when $v$ is each point of the curve, $t$ is constant and for Euclidean transform $A^TA=AA^T=Id$ and for Equi-affine transform $\det(A)=1$

Let's denote Curve $C$ after Euclidean transform as $\hat C$ and after Equi-affine transform as $\bar C$

I'm looking for any information if curve Cartan's signature is the same after Euclidean transform and after Equi-affine transform.

I mean $S(C)=S(\hat C)$ is it true? and this one $S(C)=S(\bar C)$?

sorry if something not clear, I did my best.

It's kind of thoeretical part of this question:

EDIT

My results to original question:

1. Euclidean transform keep Cartan's signature
2. Equi-affine transform change Cartan's signature

but it based on numerical results of couple of examples.

the curve that I worked with is $[cos(t),sin(t)+0.2*sin(t)^2]$ when $t\in[0,2\pi)$ in sampled in 1000 points uniformly.

Original curve

Cartan's signature

Curve after Euclidean transform $ A = \begin{pmatrix} cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{pmatrix}; b = \begin{pmatrix} 1 \\ -1 \end{pmatrix} $ when $\theta = \pi/4$

Cartan's signature after Euclidean transform

Curve after Equi-affine transform $ A = \begin{pmatrix} 1 & 1 \\ 0.5 & 1.5 \end{pmatrix}; b = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

Cartan's signature after Equi-affine transform