If $ f:\mathbb{R}^n \to \mathbb{R}^n$ is an involution, then there exists a basis $\{v_i\}$, such that $ f(v_i) = v_i $ or $ f(v_i) = -v_i$? If $ f:\mathbb{R}^n \to \mathbb{R}^n$ is an involution, then there exists a basis  $ {v_1,...,v_n } $ for $ \mathbb{R}^n$such that for each $ i $ $ f(v_i) = v_i $ or $ f(v_i) = -v_i$?
This statement seems to be true for linear involutions in particular, but I'm stuck trying to determine whether it'd be true for a general involution. I tried analysing at least the case n=1, but even there I don't have any promising ideas for a proof, but trying to construct a counterexample has failed each time, too.
Any hints?
 A: Let
$$g(x) := \begin{cases}-1&\lfloor x \rfloor \equiv 0 \pmod2\\ 1 & \lfloor x\rfloor \equiv 1\pmod2\end{cases}$$
Then
$$f(x) := x+ g(x)$$
is an involution, but for all $x\in\mathbb R: f(x) \notin\{x,-x\}$ and thus you have a (nonlinear) counter-example

For the linear case, it is indeed true, for $\mu_f | X^2-1 = (X+1)(X-1)$ and thus $\sigma(f)\subseteq\{-1,1\}$
A: If $f$ is assumed to be a linear transformation (as it seems to be) note that $$f^2-1=0$$ so the minimal polynomial of $f$ divides $X^2-1=(X-1)(X+1)$, hence $f$ is diagonalizable with possible eigenvalues $-1,1$ and the claim follows. In particular if $f$ is not a multiple of the identity, if has at least one eigenvector with eigenvalue $1$ and one eigenvector with eigenvalue $-1$.
A: If $f$ is linear involution, then it is realized by a matrix $A\in\mathbb R^{n\times n}$, i.e., $f(x)=Ax$. Involution means $f(f(x))=A^2x=x$, which means that the polynomial $p(x)=x^2-1$ annihilates $A$, and thus all its eigenvalues are $1$ or $-1$, and since  has simple roots, then $A$ is diagonalizable. So there exists a basis $v_1,\ldots,v_n$ of $\mathbb R^n$, such that
$$
f(v_i)=Av_i=\lambda_i v_i=\pm v_i, \quad i=1,\ldots,n.
$$
A: This can be done by induction on the dimension n: If $n=1$, it is clear. Now consider $dim=n$. For andy non-zero vector $v_1$, if $f(v_1) \ne -v_1$, then $f(v_1+f(v_1))=v_1+f(v_1)$ for $f^2=1$. Hence we can find non-zero vector $w_1$ such that $f(w_1)=-w_1$ or $w_1$. Now let U be the comlementary of $ \langle w_1 \rangle$, thus $V=\langle w_1 \rangle + U$. You can check $f(U)=U$. Since $dim(U)=n-1$, by induction, there exist basis of $U$ such that $f(w_i)=\pm w_i$ for $i \ge 2$. This completes the proof.
