Given any two non zero vectors $x,y$ does there exist symmetric matrix $A$ such that $y = Ax$? 
Let $x,y$ be non-zero vectors from $\mathbb{R}^n$. Is it true, that there exists symmetric matrix $A$ such that $y = Ax$.

I was reasoning the following way. Having an equation $y = Ax$ for some symmetric matrix $A$ is equivalent to being able to solve a system of $n + \frac{n^2-n}{2}$ linear equations:
$$\begin{cases} a_{11}x_1 + ... + a_{1n}x_n = y_1 \\ \vdots \\ a_{n1}x_1 + ... + a_{nn}x_n = y_n \\ a_{12} = a_{21} \\ \vdots \\ a_{n,n-1} = a_{n-1,n} \end{cases}$$
First $n$ equations are our constraints for identity $y = Ax$ to be true and the following $\frac{n^2-n}{2}$ for symmetricity of $A$. So, in our system we have $n^2$ variables and $\frac{n^2-n}{2} + n = \frac{n^2+n}{2}$ equations. As $n^2 \geq \frac{n^2+n}{2}$ such system is always solvable. That's why such matrix $A$ always exists.
This solution was marked as wrong on exam, where did I make a mistake?

It's not true that if you have more variables than equations, the system is always solvable! That's like saying a matrix with more columns than rows is always onto, which is also false.
There might be a slicker solution, but the first thing I thought of was this: Let $d = ||x||$. Then there exists a unitary matrix $U$ such that $U(x) = (d,0,0,\ldots,0)$: You can use Gram-Schmidt to create an orthonormal basis whose first vector is $x/d$, and then $U$ is the matrix whose rows are that orthonormal basis. Now, there exists a symmetric matrix A such that $A(d,0,0,\ldots,0) = U(y)$: Just set the first column of $A$ to be $U(y)/d$, and then fill in the rest of the entries to make $A$ symmetric. Now, $U^{-1}AU$ is your desired symmetric matrix sending $x$ to $y$. 
