Roots of $f(x)=f^{-1}(x),~ f:\mathbb R \rightarrow \mathbb{R}$ I find that roots of
$$\tag{1} f(x)=f^{-1}(x),$$
where $f: \mathbb{R} \rightarrow  \mathbb{R}$ are the same as those of $f(x)=x$ and $f^{-1}(x)=x$ if $f(x)$ is an increasing function. If $f(x)$ is decreasing then roots of (1) are the same as $f(x)=-x$ and $f^{-1}(x)=-x.$ Two functions: $f(x)= x^3$ and $f(x)=-x^3$ are examples.
In general, I would like to know if it is necessarily true, are there exceptions? What could be a generalization of roots of (1) in the light of these two cases.
Edit: As per the answer/response of Aditya below self inverse functions may be exceptional to this conjecture.
Edit: Can there be some generalization wherein the roots of $f(x)=f^{-1}(x)$ would be same as that of $f(x)=p(x)$ and $f^{-1}(x)=p(x)$, what could be  $p(x)$ other than $x$?

You are right only for increasing functions. For decreasing functions you are right only under an additional assumptions like oddness (see below). For instance, the function $f(x)=5-x$ is a decreasing bijection of $\mathbb R$ with $f(0)=5$ and $f(5)=0$ and thus $f(0)=f^{-1}(0)=5$ and $f(5)=f^{-1}(5)=0$.
For (necessarily strictly) increasing bijections of $\mathbb R$, however, you are right. The proof goes like this: If $f$ is strictly increasing and $f(x)=f^{-1}(x)=y$ then $f(y)=x$. But $f(x)=y$ and $f(y)=x$ together are for strictly increasing $f$ impossible in the cases $x<y$ or $y<x$. Hence, necessarily $x=y$, that is, it follows that $f(x)=x=f^{-1}(x)$.
You are also right for bijections of $\mathbb R$ which are (necessarily strictly) decreasing and odd (that is, $f(-x)=-f(x)$ for all $x$). Indeed, if $f(x)=f^{-1}(x)=y$, hence $f(y)=x$, then we have by the oddness the two equalities $f(x)=y$ and $f(-y)=-x$. These equalities imply in case $x<-y$ by the monotonicity that $y>-x$ and in case $x>-y$ by the monotonicity that $y<-x$. In both cases, one gets a contradiction by multiplying one of the inequalities with $-1$. Thus, necessarily $x=-y$, that is, it follows that $f(x)=-x=f^{-1}(x)$.
