Fix Me or Crush Me

Imagine that you have a pair of vectors ${\bf F}$ and ${\bf Z}$
$$
{\bf F}=\pmatrix{1\cr 1 \cr 0}\quad {\bf Z}=\pmatrix{0\cr 1 \cr 1}
$$
Can you construct an example of a matrix ${\bf M}$, other than the identity matrix, which leaves ${\bf F}$ fixed, in that ${\bf M}{\bf F}={\bf F}$? How many such matrices can you find? Which is the simplest? Which is the most complicated?

Can you construct an example of a matrix ${\bf N}$, other than the zero matrix, which crushes ${\bf Z}$ to the zero vector ${\bf 0}$, in that ${\bf N}{\bf Z}={\bf 0}$? How many such matrices can you find? Which is the simplest? Which is the most complicated?

Can you find a matrix which leaves ${\bf F}$ fixed and also crushes ${\bf Z}$?

Can you find any (many?) vectors fixed or crushed by the following matrices? Give examples or convincing arguments if no such vectors exist.
$$
{\bf M} = \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}, \: \begin{pmatrix} 1&2&3\\ 2&3&4\\ 3&4&5\\ \end{pmatrix}, \: \begin{pmatrix} \phantom{-}1&-2&\phantom{-}1\\ \phantom{-}1&\phantom{-}1&\phantom{-}0\\ -2&\phantom{-}1&-2\\ \end{pmatrix}
$$

You might find this Matrix Multiplication calculator helpful for testing out your ideas.

There are more matrix problems in this feature.