Constructing a circulant matrix 
Suppose two arbitrary vectors $x\in\mathbb{C}^n$ and $y\in \mathbb{C}^n$ are given. Consider determining a circulant $C\in\mathbb{C}^{n\times n}$ such that $$y = Cx$$
a.) Assume that $C$ exists for a given pair $(x,y)$, show how to construct it.
b.) When is $C$ unique for a given pair $(x,y)$
c.) When does $C$ not exist for a given pair of $(x,y)$

Note - $F$ is the unitary matrix representing the DFT and $\Lambda\in\mathbb{C}^{n\times n}$ is a diagonal matrix
Attempted solution a.) - The relationship of the eigendecompostion of $C$ to the DFT gives a simple and elegant view of the problem in the Fourier domain, i.e., the eigensdomain. We have \begin{align*} 
Cx &= y\\
F\Lambda F^H x & = y\\
\Lambda F^H x &= F^H y\\
\Lambda x &= F^H y F\\
x &= \Lambda^{-1}F^H y F
\end{align*}
Attempted solution b.) If we suppose that $A$ and $B$ are given matrices and we further suppose that $ y = Ax = Bx$ then if we show that $A = B$ does that suffice to show that $C$ is unique?
I am not sure if I am right any suggestions is greatly appreciated for a,b, and c.
 A: Note that if we define $n$ unknowns $c_k$, for $k=0,\ldots,n-1$ such that the entry $C_{i,j}$ of $n\times n$ circulant matrix $C$ is $c_k$ for $k\equiv i-j \bmod{n}$, then:
$$ C=
\begin{bmatrix}
c_0     & c_{n-1} & \dots  & c_{2} & c_{1}  \\
c_{1} & c_0    & c_{n-1} &         & c_{2}  \\
\vdots  & c_{1}& c_0    & \ddots  & \vdots   \\
c_{n-2}  &        & \ddots & \ddots  & c_{n-1}   \\
c_{n-1}  & c_{n-2} & \dots  & c_{1} & c_0 \\
\end{bmatrix}  $$
Now the requirement that $Cx=y$ for given (complex) vectors $x,y$ amounts to a system of $n$ linear equations for the $n$ unknowns $c_k$.  Simply put, we can determine the solution for $C$ by solving for the $c_k$ in this linear system, since it merely restates the condition that $Cx = y$, as desired.
A little thought shows that the actual matrix form of this linear system again involves a circulant matrix $X$ such that:
$$ X \begin{pmatrix} c_0 \\ \vdots \\ c_{n-1} \end{pmatrix} = y $$
where:
$$ X = 
\begin{bmatrix}
x_1     & x_n & \dots  & x_3 & x_2  \\
x_2 & x_1    & x_n &         & x_3  \\
\vdots  & x_2 & x_1   & \ddots  & \vdots   \\
x_{n-1}  &        & \ddots & \ddots  & x_n   \\
x_n  & x_{n-1} & \dots  & x_{2} & x_1 \\
\end{bmatrix}  $$
So the answer to part (b) amounts to the condition that $X$ is invertible, thus making the solution to $Cx=y$ unique.  It also gives us insight into how to arrange that $X$ is not invertible and hence solutions may not be unique (or may not exist, which is your part (c)).
Example
Since this approach seemingly switches the role of $C$ with circulant matrix $X$, a small worked example may help the Reader see that what is happening can be discovered by following a standard line of investigation.
Suppose $x = y = (1,2,3)^T$.  Clearly $C=I$ satisfies the requirment that $Cx=y$, but we show how the analysis above proves this is a unique solution in this particular case.  Writing out the general $3\times 3$ system of equations for entries of $C$:
$$  \begin{align*} 
c_0 x_1 + c_2 x_2 + c_1 x_3 &= y_1 \\
c_1 x_1 + c_0 x_2 + c_2 x_3 &= y_2 \\
c_2 x_1 + c_1 x_2 + c_0 x_3 &= y_3 \end{align*}  $$
Rearranging the terms so that the unknowns $c_i$ are ordered:
$$  \begin{align*} 
c_0 x_1 + c_1 x_3 + c_2 x_2 &= y_1 \\
c_0 x_2 + c_1 x_1 + c_2 x_3 &= y_2 \\
c_0 x_3 + c_1 x_2 + c_2 x_1 &= y_3 \end{align*}  $$
we arrive at the reversal of roles for $X$ and $C$:
$$ \begin{bmatrix} x_1 & x_3 & x_2 \\ x_2 & x_1 & x_3 \\ x_3 & x_2 & x_1 
\end{bmatrix} \begin{pmatrix} c_0 \\ c_1 \\ c_2 \end{pmatrix} =
\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} $$
Substitute the known values from $x,y$ and we have the specific system:
$$ \begin{bmatrix} 1 & 3 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 1 
\end{bmatrix} \begin{pmatrix} c_0 \\ c_1 \\ c_2 \end{pmatrix} =
\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} $$
The coefficient matrix $X$ here is nonsingular since $\det X = 18$, so the solution $c_0 = 1$ and $c_1 = c_2 = 0$ is unique.  Thus $C=I$ is the only possibility for a circulant matrix that makes $Cx=y$.
A: This is my professors solution that I do not really understand fully (forgive me for not writing it out):

