Prove $\forall f\in \mathbb R^{\mathbb R}\left [(f(6)=6) \to (\exists g\in \mathbb R^{\mathbb R}((g\neq i_{\mathbb R})\wedge (g\neq f)\wedge (f\circ g = g\circ f)) \right ]$
My attempt:
Let $f\in \mathbb R^{\mathbb R}$ such that $f(6)=6$, then there exists $g\in \mathbb R^{\mathbb R}$ such that $g(x)=\begin{cases}6 &x=6,7 \\ f(x)-1 & else \end{cases}$
So we have $(g\neq i_{\mathbb R})\wedge (g\neq f)\wedge (f\circ g = g\circ f)$ as desired.
(The composition is equal only at $x=6$ since there's no info for $f(x), x\neq 6$)
Also, if $f(6)\neq 6$ then any $g$ will do.
If $f \neq f \circ f$, then $f \neq id_\Bbb R$, and so picking $g = f \circ f$ is a suitable choice (and is the first thing you should be thinking about).
If $f = f \circ f$ (so $f$ is a kind of projection), there are lots of ways to proceed. For example you can have a $g$ that swaps two different fixpoints (and send their preimages to a preimage of the other fixpoint).
So if $f \neq 6$ (i.e. $f$ is not the constant function), then $f$ has another fixpoint $y$ (anything in the image of $f$ that's not $6$), so you can have $g(x) = y$ when $f(x)=6$, $g(x) = 6$ when $f(x)=y$, and $g(x) = x$ for $f(x) \notin \{6,y\}$. Then $g$ commutes with $f$ and $g \neq id_\Bbb R,6$
If $f=6$ then for any $x$, $f(g(x)) = g(f(x)) \iff 6 = g(6)$ so any function $g$ with $g(6)=6$ commutes with $f$. Then you only need to give an example of a function that's not $id_\Bbb R$, not $6$, and for which $g(6)=6$