Every free group admits a fixed-point-free involution automorphism This comes from an exercise in Rotman's "An Introduction to the Theory of Groups":

11.6) Show that a free group $F$ of rank $\geq 2$ has an automorphism $\phi$ with $\phi(\phi(w)) = w$ for all $w \in F$ and with no fixed points ($\phi(w) = w \implies w = 1$).

I started out by working out some finitely generated cases, to make it a bit easier at first, and maybe get a general idea of $\phi$ for an arbitrary free group. The case where $F$ has rank $2$ is trivial: we can swap the generators. In fact, for any even rank, the same holds. This made me feel like there was some way to pick a permutation of the generators.
Enter rank $3$. The only permutations in $S_3$ that have order $2$ are transpositions - but these obviously have a fixed point, and thus don't work. So this is where I got stuck...
I also thought about how every free group (I think - I don't recall if it's true about those of uncountable rank) is a subgroup of the free group on $2$ generators, so maybe the switch of the two would induce an automorphism on its free subgroups, but I couldn't find anyway to make it work...
Could anyone please provide a hint as per how to procede, at least in the rank 3 case? And does this result still hold for any (even uncountable) rank? Rotman's "$\geq 2$" didn't quite make it clear if he only meant finite rank for me.
Thanks in advance!
 A: Here's an order 2 automorphism $f : F \to F$ of the rank 3 free group $F=\langle x, y, z \rangle$ with no fixed points (except the identity):
\begin{align*}
f(x) &= z y^{-1} \\
f(y) &= y^{-1} \\
f(z) &= x y^{-1}
\end{align*}
The way that I got this is by using topology. The rank $3$ free group is the fundamental group of any connected, finite graph of Euler characteristic $-2$. If you can cook up a connected, finite graph $G$ with $\chi(G)=-2$, and an order $2$ self-homeomorphism $h : G \to G$ which has a fixed vertex $v$ and no fixed loops, then when you write down an isomorphism
$$\langle a,b,c \rangle \approx \pi_1(G,v)
$$
you should be able to also write down the formula for the induced fundamental group isomorphism $h_* : \pi_1(G,v) \to \pi_1(G,v)$ which will be of order 2 (because $h$ has order 2) and has no fixed elements (because $h$ has no fixed loops).
I used the graph $G$ with two vertices $p,q$ and three edges $\alpha,\beta,\gamma,\delta$ each oriented with intial vertex $p$ and terminal vertex $q$. This graph has an order $2$ self-homeomorphism $h$ fixing each of $p$ and $q$, defined by
\begin{align*}
h(\alpha) &= \gamma\\
h(\gamma) &= \alpha\\
h(\beta) &= \delta\\
h(\delta) &= \beta
\end{align*}
The isomorphism $\langle a, b, c \rangle \mapsto \pi_1(G,v)$ is given by
\begin{align*}
a &\approx \alpha \, \bar\delta \\
b &\approx \beta \, \bar \delta \\
c &\approx \gamma \, \bar \delta
\end{align*}
and deriving the formula for $h_*$ is now just a computation. For example
\begin{align*}
h_*(a) &= [h(\alpha \, \bar \delta)] \\
 &= [\gamma \, \bar  \beta] \\
 &= [\gamma \, \bar  \delta \,  \delta \, \bar  \beta] \\
 &= [\gamma \, \bar  \delta] \, [\beta \, \bar  \delta]^{-1} \\
 &= z  \, y^{-1}
\end{align*}

It should not be hard to generalize this to any free group of odd finite rank.
And for groups of any infinite rank, just subdivide the free basis into subsets of size $2$ and do what you did for even ranks.
A: Promoting my comment to an answer:  let $\phi$ be the automorphism sending each generator of $F$ to its inverse.  This clearly has order 2 (for free groups of arbitrary rank $\geq 1$), and since it sends each reduced word $x_{i_1}^{e_1} \cdots x_{i_n}^{e_n}$ to the reduced word $x_{i_1}^{-e_1} \cdots x_{i_n}^{-e_n}$, its only fixed point is the identity.
