Let $O(d)$ denote the orthogonal group for $\mathbb{R}^d$ with the standard Euclidean inner-product. A nautral question is to find all possible functions $f: \mathbb{R}^d \to \mathbb{R}$ that are invariant under the group action in the sense that $f(Rx) = f(x)$ for all $x \in \mathbb{R}^d$ and $R \in O(d)$. Since $O(d)$ acts transitively, we can fix our favorite unit vector $u \in \mathbb{R}^d$ and pick $R \in O(d)$ such that $x = R \vert x \vert u$ to deduce that $$ f(x) = f(R \vert x \vert u) = f(\vert x \vert u), $$ which means that if we set $\varphi : \mathbb{R} \to \mathbb{R}$ via $\varphi(r) = f(r u)$ then we arrive at the identity $$ f(x) = \varphi(\vert x\vert). $$ In other words, the most general form of such an invariant function is a "radial function" of the form $x \mapsto \varphi(\vert x \vert)$.
We can play a similar game if we shift the codomain to be $\mathbb{R}^d$ rather than $\mathbb{R}$. In this case we can look for equivariant functions $f: \mathbb{R}^d \to \mathbb{R}^d$ such that $f(Rx) = R f(x)$ for all $x \in \mathbb{R}^d$ and $R \in O(d)$. As above, we find $$ f(x) = R f(\vert x \vert u). $$ By examining the subgroup of $O(d)$ that fixes $u$ we can see that the map $u \mapsto f(\lambda u)$ must be parallel to $u$ for $\lambda \ge 0$, or in other words, $f(\lambda u) = \varphi(\lambda) u$. Plugging this back in shows that $$ f(x) = R \varphi(\vert x\vert) u = \frac{\varphi(\vert x\vert)}{\vert x\vert} R \vert x \vert u = \frac{\varphi(\vert x\vert)}{\vert x\vert} x =: \psi(\vert x\vert) x, $$ and we now have the most general form of such equivariant vector fields.
There's no reason to stick to one argument, though. We can fix an integer $n \ge 2$ and look for invariant functions $f: (\mathbb{R}^d)^n \to \mathbb{R}$ such that $f(Rx_1,\dotsc, R x_n) = f(x_1,\dotsc, x_n)$ for all $x_1,\dotsc,x_n \in \mathbb{R}^d$ and $R \in O(d)$. Here the algebra gets a bit more involved. For instance, if we want all polynomial functions satisfying this, then the classification of such functions is provided by the fundamental theorem of the orthogonal group, which says that all such polynomials are generated by the $n^2$ inner-products $(x_i,x_j)$ for $1\le i,j \le n$. If we allow for the more general case and drop the requirement that $f$ be polynomial, then the fact that we have more than one argument seems to preclude using an argument as simple as the one above: $$ f(x_1,\dotsc,x_n) = f(R \vert x_1 \vert u, R R^{-1} x_2, \dotsc, R R^{-1} x_n) = f( \vert x_1 \vert u, R^{-1} x_2, \dotsc, R^{-1} x_n), $$ but then I'm not sure how to deal with the last term as it doesn't seem to have any good invariant properties to exploit. I would guess, though, that the general form should be a function of just the $n^2$ inner-products.
My real question, though, is about what happens when we examine the multi-variate version for the equivariant problem. Can we classify all equivariant functions $f: (\mathbb{R}^d)^n \to \mathbb{R}^d$ such that $f(Rx_1,\dotsc, R x_n) = R f(x_1,\dotsc, x_n)$ for all $x_1,\dotsc,x_n \in \mathbb{R}^d$ and $R \in O(d)$? I'm perfectly happy here with arbitrary functions, i.e. they don't need to be of polynomial type, and I would guess this makes the classification less of an algebraic mess. Such functions certainly exists. For instance, $$ f(x_1,\dotsc,x_n) = \sum_{i=1}^n \varphi_i((x_1,x_1),(x_1,x_2),\dotsc,(x_n,x_n)) x_i $$ does the trick, i.e. we can take linear combinations of scalar functions of the $n^2$ inner-products multiplied by the vectors in the tuple. Perhaps a version of the fundamental theorem is known for such equivariant maps, but I was unable to find it. Any assistance in either the general classification or finding the appropriate reference would be greatly appreciated.