Describe all matrices similar to a certain matrix. Math people:
I assigned this problem as homework to my students (from Strang's "Linear Algebra and its Applications", 4th edition): 

Describe in words all matrices that are similar to $$\begin{bmatrix}1& 0\\ 0& -1\end{bmatrix}$$ and find two of them.

Square matrices $A$ and $B$ are defined to be "similar" if there exists square invertible $M$ with $A = M^{-1}BM$ (or vice versa, since this is an equivalence relation).  The answer to the problem is not in the text, and I am embarrassed to admit I am having trouble solving it.  The problem looked easy when I first saw it.
The given matrix induces a reflection in the $x_2$-coordinate, but I don't see how the geometry helps.  A similar matrix has to have the same eigenvalues, trace, and determinant, so its trace is $0$ and its determinant is $-1$.  I spent a fair amount of time on it, with little progress, and I can spend my time more productively.  This problem is #2 in the problem set, which suggests that maybe there is an easy solution.
I would settle for a hint that leads me to a solution.    
EDIT: Thanks to Thomas (?) for rendering my matrix in $\LaTeX$.
Stefan (STack Exchange FAN)

Make a picture, your matrix mirrors the $e_2$ vector and doesn't change anything at the $e_1$ vector. The matrix is in the orthogonal group but not in the special orthogonal group. Show that every matrix 
$$\begin{pmatrix} \cos(\alpha) & \sin(\alpha) \\ \sin(\alpha) & -\cos(\alpha)\\ \end{pmatrix} $$ 
make the same.
Those are the nicest matrix which can happen to you but there are some more (those matrices appear when $M$ itself is in the orthogonal group. 
When $M$ is not in the orthogonal group it still won't change the eigenvalues (I am not sure if you already know waht eigenvalues are), $\lambda$ is an eigenvalue to a vector $v\neq 0$ if 
$$ A \cdot v=\lambda v$$ 
which means the vector is only enlarged or made smaller through the matrix, but not rotated or something like that. As $A$ has the eigenvalues $1$ and $-1$  you will always find vectors $v_1,v_2$ such that 
$$ B \cdot v_1= v_1$$ and 
$$ B\cdot v_2= -v_2.$$
So those matrices won't change one vector and the other one is "turned around".
The eigenvectors of the matrix:
$$ \begin{pmatrix} a & b \\c & d \\ \end{pmatrix}^{-1}\cdot \begin{pmatrix} 1 & 0 \\ 0 & -1\\ \end{pmatrix} \cdot \begin{pmatrix} a & b \\c & d \\ \end{pmatrix}$$
are
$$\begin{pmatrix} \frac{a}{c} \\ 1 \end{pmatrix} \qquad \begin{pmatrix} \frac{b}{d} \\1 \end{pmatrix} $$ 
when $c$ and $d$ are not zero,
