If $A$ is an orthogonal real matrix that commutes with all orthogonal matrices then $A$ is a scalar multiple of identity matrix.
I was able to prove the previous part of the question which is the following: If $A$ is a "symmetric" real matrix that commutes with all orthogonal matrices then $A$ is a scalar multiple of identity matrix.
If $A$ commutes with every orthogonal matrix, then it commutes with every Householder matrix. That is, for every unit-vector $x$, we have $A(I - 2xx^T) = (I - 2xx^TA)$. That is, for every unit vector $x$, we have $$ A(I - 2xx^T) = (I - 2xx^TA) \iff\\ A - 2(Ax)x^T = A - 2x(x^TA) \iff\\ 2(Ax)x^T = 2x(x^TA) \iff\\ (Ax)x^T = x(x^TA) $$ Note that the column-space of the first matrix is spanned by $Ax$, and the column-space of the second matrix is spanned by $x$. So, in order for these matrices to be the same, $Ax$ needs to be a multiple of $x$ for every possible $x$. That is, every unit vector $x$ is an eigenvector of $A$.
This can only happen if $A$ is a multiple of the identity; the conclusion follows.
Once we have $Axx^T = xx^TA$, another option is to use explicit matrix computation. If we take $x$ to be one of the standard basis vectors $e_1,e_2,\dots,e_n$, then we find that $$ Axx^T = xx^TA \implies (Ae_j)e_j^T = e_j(e_j^TA) \implies\\ \pmatrix{a_{1j}\\ \vdots \\ a_{nj}} e_j^T = e_j \pmatrix{a_{j1} & \cdots & a_{jn}} \implies\\ \pmatrix{0&&a_{1j}&&0\\ \vdots & \cdots & \vdots & \cdots & \vdots \\ 0&&a_{nj}&&0} = \pmatrix{0 & \cdots & 0\\ &\vdots\\ a_{j1} & \cdots & a_{jn}\\ & \vdots \\ 0 & \cdots & 0} $$ In order for this to hold for any $j$ from $1$ to $n$, it must be the case that $a_{ij} = 0$ whenever $i \neq j$. That is, $A$ must be a diagonal matrix.
In order to deduce that the diagonal entries of $A$ are the same, it suffices to check what happens when we plug in $x = e_i + e_j$ for $i \neq j$.
