Let $x$ be an n-dimensional vector, and $M$ be an $ n \times n$ matrix. Then, $M$ has at most $n$ unitary eigenvectors, i.e., vectors such that $Mx = x$.
Now, let $\phi: \mathbb{R} \rightarrow \mathbb{R}$ be any continuous function that, when applied on a vector, acts on every entry of it. Is it possible to have a $\phi$ such that the map $\phi(Mx)$ (which is now not linear anymore) has more eigenvectors? i.e., vectors such that $\phi(Mx) = x$
If yes, given a set of vectors $(x_1,...,x_m)$ and a function $\phi$, is it possible to find a matrix $M$ such that $\phi(Mx) = x_i$ holds, for every $i$?
The linear map (on $\mathbb{R}$) $M: x\mapsto 2x$ has a $1$-eigenspace of dimension $0$. Define $\phi:\mathbb{R}\to\mathbb{R}$ by $x\mapsto\frac{1}{2}x$; this is continuous.
Then the set (space in this case) of vectors with $\phi(Mx)=x$ has dimension $1$ so there are many more of them.