I think, one can argue as follows.
Let $D\subseteq\text{Sym}$ be the diagonal matrices. Since $\exp:D\to D^+$ and $\exp:\text{Sym}\to\text{Sym}^+$ are compatible diffeomorphisms it suffices to answer the analogous problem for $D\subseteq\text{Sym}$.
For $m=0,\ldots,n-1$ let $c_m:D\to D:(x_i)\mapsto(x_i^m)$. These are certainly $S_n$-covariants. It is well known that they freely generate the space of all polynomial $S_n$-covariants as a module over the ring of all polynomial invariants. I claim that this also holds for smooth functions. In other words, for every smooth covariant $a$ there are smooth invariants $f_0,\ldots,f_{n-1}$ with $$ a=\sum_{m=0}^{n-1}f_mc_m $$ There is probably a general theorem but here is an ad hoc argument which independently shows that the $c_m$ form a basis. Let $a=(a_1,\ldots,a_n)$ be the components. Then we have to solve $$ a_i=\sum_{m=0}^{n-1}f_m x_i^m $$ This is a linear system of equations for the $f_m$ with the Vandermonde matrix as coefficients. So we can uniquely solve it. One gets $f_m=\frac{\tilde f_m}{V}$ where $\tilde f_m$ is smooth and $V$ is the Vandermonde determinant. The equivariance of $a$ implies that $\tilde f_m$ vanishes where $V$ vanishes so $f_m$ is a smooth function.
Glaeser's theorem shows that $f_i(x)$ can be extended to a smooth $O(n)$-invariant $F_i(X)$ on $\text{Sym}$. Thus $$ A(X)=\sum_{m=0}^{n-1}F_m(X)X^m $$ is a smooth extension of $a$.