Stabiliser of a curve under Affine Transformations? Let $\gamma$ be a curve in the plane, and let $\text{Im}(\gamma) \subset \mathbb{R}^{2}$ be its image in the plane. Is it possible to completely specify the affine transformations of $\mathbb{R}^{2}$ which send $\text{Im}(\gamma)$ to itself?   

For example, if $\gamma(t)=(t,t^{2})$, then the linear maps $(x,y) \to (\lambda x,\lambda^{2} y)$ are (I think) the whole collection of such maps. If $\gamma$ is a circle, then the orthogonal group does the trick. I would expect that "usually" the group of stabilising actions would be trivial, but I'm not sure how to tell.
These linear maps $(x,y)→(λx,λ^2y)$ is called parabolic rescaling and is crucial in proving decoupling theorem and Vindogronov main theorem.

Now assume we have a curve defined by $f(x,y)=0$. the CORRECT ANSWER OF THIS PROBLEM is those curve that do not have "error term". I will explain what I said in the following.

1.If $f(x,y)=0$ is a plane $V$. then we know the affine map send $image(\gamma)$ to itself is just a affine map make $V$ to be a invariance subspace.

2.if $f(x,y)=0$ is a quadratic curve. Then it is easy to see $f(x,y)=0$ could not to be a elliptic. now change a basis we can assume the new basis is the eigenvalue of affine map $A$. So $Ax=\lambda_1x, Ay=\lambda_2 y$.Now I claim $f(x,y)$ should be have only two terms, every unknown quantity a term. for example $x^2-y^2,x^2-y,x^2-y$. These is just derive by $\frac{f(\lambda_1^kx,\lambda_2^ky)}{\lambda_1^{w_1k}\lambda_2^{w_2k}}= f(x,y),\forall k\in N^*$.

3.For the cubic and higher polynomial the result is similar.
These linear maps $(x,y)→(λx,λ^2y)$ is called parabolic rescaling and is crucial in proving decoupling theorem and Vindogronov main theorem.

Now assume we have a curve defined by $f(x,y)=0$. the CORRECT ANSWER OF THIS PROBLEM is those curve that do not have "error term". I will explain what I said in the following.

1.If $f(x,y)=0$ is a plane $V$. then we know the affine map send $image(\gamma)$ to itself is just a affine map make $V$ to be a invariance subspace.

2.if $f(x,y)=0$ is a quadratic curve. Then it is easy to see $f(x,y)=0$ could not to be a elliptic. now change a basis we can assume the new basis is the eigenvalue of affine map $A$. So $Ax=\lambda_1x, Ay=\lambda_2 y$.Now I claim $f(x,y)$ should be have only two terms, every unknown quantity a term. for example $x^2-y^2,x^2-y,x^2-y$. These is just derive by $\frac{f(\lambda_1^kx,\lambda_2^ky)}{\lambda_1^{w_1k}\lambda_2^{w_2k}}= f(x,y),\forall k\in N^*$.

3.For the cubic and higher polynomial the result is similar.