given any two non zero vectors $x$ and $y$ in $\mathbb{R}^n$, does there exist an invertible matrix $A$ such that $Ax=y$? 
Given any two non zero vectors $x$ and $y$ in $\mathbb{R}^n$, does there exist an invertible matrix $A$ such that $Ax=y$?

I was examining the orbits of set $\mathbb{R^n}$ under the action of the group $GL_n(R)$ by left multiplication i.e $O_x=\{Ax| A\in GL_n(R)\}$ and $x\in \mathbb{R^n}$
Since $O_0=\{0\}$, I thought there is only one more orbit of nonzero vectors. This can be shown if my above question holds true.
Thanks for the help!!
 A: i will make copper.hats answer more concrete by constructing the matrices for the case $n = 5.$ 
first i will deal with the case the first component of both $x$ and $y$ is $1.$
the matrix $Y = \left( \begin{array}{lllll} 1 & 0 & 0 & 0 & 0\cr y_2 & 1 & 0 & 0 & 0\cr y_3 & 0 & 1 & 0 & 0\cr y_4 & 0 & 0 & 1 & 0\cr y_5 & 0 & 0 & 0 & 1\end{array} \right)$
and its inverse $Y^{-1} =\left( \begin{array}{lllll} 1 & 0 & 0 & 0 & 0\cr -y_2 & 1 & 0 & 0 & 0\cr -y_3 & 0 & 1 & 0 & 0\cr -y_4 & 0 & 0 & 1 & 0\cr -y_5 & 0 & 0 & 0 & 1\end{array} \right)$ have the property $Ye_1 = y$ and $Y^{-1}y = e_1.$ in the same way form the matrices $X$ and its inverse $X^{-1}.$
now, the matrix $YX^{-1}$ will send $x$ to $y$ for $YX^{-1}x = Ye_1 = y.$ 
if the first components of $y$ or $x$ is not $1$ use permutation matrices and scaling to make the first component of both to $1.$
A: Here is a brute force approach to show existence.
Given a non-zero vector $v_1$ add vectors so that $v_1,...,v_n$ form a basis and let $V$ be the resulting matrix, that is
$V=\begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}$. Then $V e_1 = v$
($e_1$ is the vector $(1,0,\cdots,0)^T$). Note that $V$ is invertible.
So, given $x$, find $V_1$ such that $V_1 e_1 = x$, then $e_1 = V_1^{-1} x$.
Now, given $y$, find $V_2$ such that $V_2 e_1 = y$. then $V_2 V_1^{-1} x = y$, so letting $A = V_2 V_1^{-1}$ will do the trick.
Aside: If $\|v_1\| = 1$ above, the basis can be chosen to be orthonormal, and so the resulting $V$ will be orthogonal (unitary).
Hence we can find an orthogonal (unitary) $A$ such that ${ \|y\| \over \|x\| } Ax = y$.
