This is a followup from a question I asked on math.SE, which received a helpful answer but unfortunately not a complete one. $\def\Sym{\mathrm{Sym}_{n\times n}}$ $\def\s{\mathrm{Sym}}\def\sp{\s^+}$Let $\sp \subset GL(n,\mathbb R)$ denote the space of symmetric positive-definite $n \times n$ matrices. (It might be more fruitful to think of the full subspace of symmetric matrices, since it turns all the group actions below in to representations.)

I am interested in functions $A : \sp \to \sp$ that are equivariant under the natural conjugation action of $O(n)$; i.e. such that$$A(R^T X R) = R^T A(X) R$$ for all $X \in \sp, R \in O(n,\mathbb R)$.

Since we can diagonalize any $X$, we know that such an $A$ is determined by its restriction to diagonal matrices, which gives a function $a : (0,\infty)^n \to (0,\infty)^n$ which is equivariant under the natural permutation action of the symmetric group $S_n$. Conversely, any such $a$ can be extended uniquely to an equivariant $A$. Thus we can specify an $A$ by just declaring what it does to eigenvalues.

Question: If we know $a$ is smooth, can we conclude $A$ is smooth?

In the analogous problem for $O(n)$-invariant scalars $F : \sp \to \mathbb R$ (which reduce to symmetric functions $f : (0,\infty)^n \to \mathbb R$ of the eigenvalues), we can solve this problem using Glaeser's "differentiable Newton's theorem" - we get that a smooth symmetric function of the eigenvalues is a smooth function of the symmetric matrix invariants, which are in turn smooth functions of the matrix itself. The key is that $S_n$-invariant polynomials of the eigenvalues and $O(n)$-invariant polynomials of the matrices are the same thing.

Since I couldn't find any similar work on equivariant maps (please relieve me of my ignorance!), my only thought was to use something like this theorem of Schwarz to study the scalar $\tilde A : \sp \times \sp\to \mathbb R$ defined by $$\tilde A(X,Y) = \langle A(X), Y\rangle,$$ which is invariant under the action $\rho_R(X,Y) = (R^T X R, R^T Y R)$. We can define an $S_n$-invariant $\tilde a : (0,\infty)^n \times (0,\infty)^n \to \mathbb R$ similarly; but unfortunately it seems to me that there is no obvious relation between $\tilde a$ and $\tilde A$ - we only have $a(\lambda(X)) \cdot \lambda(Y) = \langle A(X), Y \rangle$ when $X,Y$ have the same eigenvectors.

Any pointers would be great - this is a tangent from my usual research, so there is probably a whole body of relevant work I'm unaware of.