Let $M_{i,j}$ be a matrix of the standard basis of size $n>1$ (the coefficients are $1$ at position $i,j$ and $0$ elsewhere)
What is the least number of orthogonal matrices required so that $M_{i,j}$ can be expressed as a linear combination of those ?
I was able to find the answer thanks to @TedShifrin.
The answer is two.
Let $M_{i,j}$ be the matrix with coefficients $1$ at $i,j$, $0$ elsewhere.
Take two matrices $A,B=((a_{k,l}))_{1\le k,l\le n},((b_{k,l}))_{1\le k,l\le n}$ such that :
$a_{i,j}=b_{i,j}=1$
On each other line/column, there is exactly one $1$ for $A$ (resp. one $-1$ for $B$)
That way, $A,B$ are orthogonal and $\frac{1}2(A+B)=M_{i,j}$
Example for $n=4,i=2,j=3$ :
$M_{i,j}=\begin{bmatrix} 0&0&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$
Let $A=\begin{bmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1 \end{bmatrix},B=\begin{bmatrix} -1&0&0&0\\0&0&1&0\\0&-1&0&0\\0&0&0&-1 \end{bmatrix}$
Then $A+B=2M_{i,j}$