How to represent $\mathbb{R}^n$ as a matrix? Is there a basis in which Hadamard product in $\mathbb R^n$ is isomorphic to the matrix product, and on the matrix main diagonal all elements be equal to the mean value of the elements of the vector in $\mathbb R^n$?
For instance, in 2-dimensional case we have
$(a,b)\to\left(
\begin{array}{cc}
 \frac{a+b}{2} & \frac{b-a}{2} \\
 \frac{b-a}{2} & \frac{a+b}{2} \\
\end{array}
\right)$
What would be in the $n$-dimensional case?
 A: I don't know if this is possible for all $n$, but it is possible for infinitely many $n$. In particular, if there exists a Hadamard matrix (an $n \times n$ matrix $H$ with $\pm 1$ entries satisfying $H^TH = nI$) of size $n$, then it follows that a suitable representation exists.
Indeed: suppose that $H$ is a Hadamard matrix of size $n$. The matrix $Q = H/\sqrt{n}$ is orthogonal. It follows that the map $f:v \mapsto Q\operatorname{diag}(v) Q^T$ is such that $f(v \odot w) = f(v)f(w)$. To see that the diagonal entries of $f(v)$ are equal to the average of $v$, let $q_1,\dots,q_n$ denote the columns of $Q$. Each vector $q_i$ has entries $\pm 1/\sqrt{n}$, which means that the diagonal entries of $q_iq_i^T$ are equal to $1/n$. It follows that the diagonal entries of
$$
f(v) = \pmatrix{q_1 & \cdots & q_n} \pmatrix{v_1 \\ &\ddots \\ &&v_n} \pmatrix{q_1^T\\ \vdots \\ q_n^T} = v_1\,q_1q_1^T + \cdots + v_n\  q_nq_n^T
$$
are equal to $(v_1 + \cdots + v_n)/n$.

The answer to the corresponding question over $\Bbb C$, on the other hand, is yes. In particular, if $U$ denotes the unitary DFT matrix and $U^*$ denotes its conjugate transpose, then the map defined by
$$
f(v) = U \operatorname{diag}(v) U^*
$$
takes vectors to circulant matrices and has all the desired properties.
For the case of $n=2$, this turns out to be the map $(a,b)\to\left(
\begin{array}{cc}
 \frac{a+b}{2} & \frac{b-a}{2} \\
 \frac{b-a}{2} & \frac{a+b}{2} \\
\end{array}
\right)$ that you mentioned, but for larger $n$ it maps some real vectors to matrices with (non-real) complex entries.
A: This is the Mathematica code implementation of the both methods by @BenGrossman:
V := {1, 2, 3, 4};
n := Length[V];
FourierMatrix[n] . DiagonalMatrix[V] . ConjugateTranspose[
        FourierMatrix[n]] // FullSimplify // ExpandAll // MatrixForm // Print; 
HadamardMatrix[n] . DiagonalMatrix[V] . ConjugateTranspose[
        HadamardMatrix[n]] // FullSimplify // ExpandAll // MatrixForm // Print;

For $n=2$, the both methods return the same result, the one from the question.
UPDATE
Defining
H(n):=PadLeft[Array[JacobiSymbol[#2 - #1, n - 1] &, {n, n} - 1] - 
  IdentityMatrix[n - 1], {n, n}, 1]

and inserting H(n) intead of HadamardMatrix(n) will make the second formula to work with any n which is a multiple of $4$ and a prime plus one at the same time, and not only with powers of $2$.
