A way to reframe the definitions: matrices $A,B: \Bbb R^n \to \Bbb R^m$ are equivalent if there exists bases $\mathcal B_1, \mathcal B_2$ of $\Bbb R^n,\Bbb R^m$ such that $B$ is the matrix of $A$ with respect to these new bases.
However, in the case of square matrices, we might say that choosing different bases for the first and second space loses too much structure from the original map $A$. For instance, take
$$
A = \pmatrix{0&1\\1&0}
$$
(the reflection through $y=x$). This map is equivalent to the identity map. However, $A$ is fundamentally different from $I$ in that $A$ changes the space, and $I$ doesn't. The problem here is that if we can change our perspective in the output space, we can make it seem as though $A$ didn't do anything at all; we can take the image of our input basis to be our output basis.
In order to get an idea of how $A$ changes a space, we restrict ourselves to similarity. That is, we impose the constraint that the basis we choose for the output space should be the same as what we choose for the input space; this basis provides us with a fixed frame of reference. A useful choice of basis, in this instance, is $\mathcal B = \{(1,1),(1,-1)\}$. Relative to this basis, we find that
$$
[A]_{\mathcal B} = B = \pmatrix{1&0\\0&-1}
$$
we see that the similar matrix $B$ is clearly a reflection since it flips the sign of the $y$-coordinate.
One notable property is that if $A$ is similar to $B$, then $A^n$ is similar to $B^n$ for any $n$.
