For $1\leq k,j \leq n$ and $a=(a_1,\ldots,a_n)\in {\mathbb R}^n$, denote by $s_{k,j}(a)$ the $k$-th symmetric polynomial in the $n-1$ variables obtained when $a_j$ is removed from $a_1,\ldots,a_n$. For example,

$$ s_{2,3}(a_1,a_2,a_3,a_4,a_5)= a_1(a_2+a_4+a_5)+a_2(a_4+a_5)+a_4a_5 $$

Next, define a map $f_{n,k}:{\mathbb R}^n \to {\mathbb R}^n$ by

$$ f_{n,k}(a)=(s_{k,1}(a),s_{k,2}(a),\ldots,s_{k,n}(a)) $$

Note that $f_{n,1}$ is a linear injective map on ${\mathbb R}^n$, and at the other extreme, using multiplication instead of addition, $f_{n,n}$ is injective on $({\mathbb R}^+)^n$. Also, for even $k\gt 1$, $f_{n,k}$ is not injective on ${\mathbb R}^n$ as it is invariant by multiplication by a $k$-th root of unity.

Question: For $1\lt k \lt n$, is $f_{n,k}$ injective on $({\mathbb R}^+)^n$ ?

It is easy to see that $a_1,a_2,\ldots,a_n$ are algebraic on ${\mathbb Q}(s_{k,1}(a),s_{k,2}(a),\ldots,s_{k,n}(a))$, but computing the minimal polynomials in the general case seems hopeless.

Cross-posted from MSE with a few minor modifications.