Are $f(x)=x$ and $f(x)=-x$ the only odd bijective involutions from $\mathbb{R}$ to $\mathbb{R}$? This is motivated by a question that was posted last night, but was deleted (I think by the author) before any answers to it appeared. I don't think it has been re-posted since then.

What bijective functions $f$  from $\mathbb{R} \rightarrow \mathbb{R}$ satisfy $$f(x)=\frac{f^{-1}(x)-f^{-1}(-x)}{2}?$$

I first spotted that the RHS was the "odd part" of the inverse function, making $f$ an odd function. But this implies that $f^{-1}$ is odd, making the RHS equal to $f^{-1}$. So I end up with the following information: 


*

*$f$ is odd

*$f(x)=f^{-1}(x)$.


This is where I got stuck. The only $\mathbb{R} \rightarrow \mathbb{R}$ functions symmetric with respect to both the origin and the line $y=x$ that I can visualize are $f(x)=x$ and $f(x)=-x$, but I am not sure if they are indeed the only ones. 

How I can prove/disprove that $f(x)=x$ and $f(x)=-x$ are the only odd, bijective $\mathbb{R} \rightarrow \mathbb{R}$ involutions?

 A: As noted in the comments, there are a lot of possibilities if $f$ is not assumed continuous.
Under the assumption that $f$ is continuous, however, $f(x)=+x$ and $f(x)=-x$ are the only odd involutions of $\mathbb{R}$. Let $f$ be such a function.
As $f$ is a continuous bijection of $\mathbb{R}$, it is strictly monotonic. Since $g(x)=-f(x)$ is also odd, and also an involution since $g(g(x))=-f(-f(x))=x$, we may (and will) assume that $f$ is strictly increasing. We have to show that $f(x)=x$ for all $x$.
Let $E^+=\{x  : f(x)>x \}$, $E^-=\{x \, : f(x)<x\}$. These sets are open by continuity of $f$, hence (at most countable) union of disjoint open intervals. Since $f$ is odd, $f(0)=0$ so $0\notin E^+\cup E^-$.
Let $(a,b)$ be such an interval, we first assume this is in $E^+$, let us have a look at the case where $(a,b)\subset E^+$. Since $0\notin E^+$, $a$ or $b$ has to be finite. 
Say this is $a$. Since $a\notin E^+$, and $f$ continuous, $f(a)=a$. By continuity, and because $f$ is increasing, $f(a+\epsilon)\in (a,b)\subset E^+$ for small enough $\epsilon>0$. So by definition of $E^+$, 
$$f(f(a+\epsilon))>f(a+\epsilon)>a+\epsilon.$$
Since $f$ is an involution, we get a contradiction. The four other cases ($b$ is finite but $a$ is not, or $(a,b)\subset E^-$, $a$ or $b$ is finite, are handled similarly.
