I have a conjecture inspired by the following observation. If $f:\mathbb{R}\to \mathbb{R}$ is a continuous bijective function that satisfies $f(x)+f^{-1}(x)=2x$ and has a fixed point, then $f(x)=x$. This can be proved quite easily, though not trivial.
I conjecture that if we omit the assumption that $f$ has a fixed point, then $f(x)=x+d$ for some $d\in \mathbb{R}$. More generally, I think that if $f:\mathbb{R}^n \to \mathbb{R}^n$ is continuous bijection which satisfies $f(x)+f^{-1}(x) = \Lambda x$ for some bijective linear map $\Lambda:\mathbb{R}^n \to \mathbb{R}^n$, then $f$ must be linear.
I first want to tackle a seemingly easier problem: does there exist a bijective linear map $S$ with $S+S^{-1}=\Lambda$? Unfortunately I could not handle this problem, so I would really appreciate it if you could give me some suggestions or a counterexample.
Answer
Okay, here's a proof sketch for your first conjecture.
Let $f:\mathbb R \to \mathbb R$ be a continuous bijection, therefore monotonic. Suppose $f(x) + f^{-1}(x) = 2x$. Then for all $a\in \mathbb R$ and $n\in\mathbb Z$, $f$ also satisfies $f^n(a) = a+nd_a$ where $d_a = f(a) - a$, by induction on $n$. You've already handled the case where some $d_a=0$, so assume that all $d_a\neq0$. Then we've described the behavior of $f$ on a bunch of lattices in $\mathbb R$.
Now, if for some $a_1, a_2 \in \mathbb R$ we have $d_{a_1}\neq d_{a_2}$, then the two lattices don't interleave properly at large values of $n$ (starting somewhere around $(a_2-a_1)/(d_{a_1}-d_{a_2})$, I think). So $f$ is not monotonic, a contradiction. Therefore $d_a$ doesn't depend on $a$, and $f(x)=x+d_0$ for all $x$.
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