The matrices A and B are similar if there is some nonsingular matrix P such that PA = BP. Nonzingular means invertible, or at least invertible in the fractions of R, so an equivalent definition asserts PA/P = B.
Set P to the identity matrix, and every matrix is similar to itself.
If A and B are similar as above, let Q be P inverse, and rewrite the equation as A = QB/Q, hence the relation is symmetric.
The last property is transitivity. If PA/P = B and QB/Q = C, then QPA/P/Q = C, and A and C are similar via the matrix PQ. Therefore "similar to" forms an equivalence relation on matrices.
Let's try to provide a geometric interpretation. Every matrix corresponds to a linear function from n space into n space. A nonsingular matrix implements a change of basis. In other words, the coordinates are moved about in a manner that can be reversed, i.e. they aren't squashed into fewer dimensions. Therefore PA/P moves n space about, applies the function associated with A, and pulls space back again. Combine these into one matrix, B, to describe the composite function. We are looking at the same transformation, from another point of view.
The rank of a matrix is the dimension of its range. If A has rank r then PA cannot have a larger rank, nor can PA/P. Thus rank(B) ≤ rank(A). However, similarity is a symmetric relation, so rank(A) ≤ rank(B), and rank(A) = rank(B). Similar matrices have the same rank.
Beyond this, similar matrices have the same determinant. Write det(P)det(A) = det(B)det(P), and remember that det(P) is nonzero.
Let's close this section by defining a couple of adverbs. The matrices A and B are orthogonally similar if P is orthogonal, and unitarily similar if P is unitary. Since the inverse of a unitary matrix is unitary, and the product of unitary matrices is unitary, unitarily similar matrices fall into equivalence classes. We'll see, later on, that one can always choose an upper triangular matrix to represent each class.