Let $X\subset\mathbb{A}^n_{\mathbb{C}}$ isomorphic to $\mathbb{A}^1_{\mathbb{C}}$.
(a) Assume $X\subset \{x_n=0\}\subset\mathbb{A}^n_{\mathbb{C}}$. Prove there exists an automorphism $f$ of $\mathbb{A}^n_{\mathbb{C}}$ such that $f(X)$ is a line.
(b) Don't assume part (a). It is known that for $n\geq4$ that after a change of variables the projection $p:\mathbb{A}^n_{\mathbb{C}}\to\mathbb{A}^{n-1}_{\mathbb{C}}$ induces an isomorphism of $X$ and $p(X)\subset\mathbb{A}^{n-1}_{\mathbb{C}}$. Use this to prove that there is an automorphism $f$ of $\mathbb{A}^n_{\mathbb{C}}$ such that $X$ is a line.
(c) Assume $n=3$. Is there always an automorphism $f$ of $\mathbb{A}^3_{\mathbb{C}}$ such that $f(X)$ is a line?
This a problem in my homework of Introduction to Algebraic Geometry, and I really have no clue how to do this.
We tried some things with my classmates but they led us to nothing. We tried using the isomorphism of X with $\mathbb{A}^1$ (let's call it $g$) to define $f(x_1,\ldots,x_n)=(x,g(x_1,\ldots,x_n)+x_n)$ or similar variations of this. Unfortunately, this attempts didn't work. Geometrically, we are breaking $\mathbb{A}^n$ into the hyperplane where $X$ lives and the orthogonal part (like a free dimension to use). In the hyperplane ($\mathbb{A}^{n-1})$ we are trying to "strecth" the curve to an axis while leaving everything unchanged. We believe that we have to use the isomorphism $g$ in the following manner: Extend $g$ to the entire $\mathbb{A}^{n-1}$. Then, use the "extra dimension" to correct bijectivity. That is all we have.