I know that $I$ has only one similar matrix, so does $O$, and I also know that any matrix with finite dimensions has a Jordan form. But my question is:
How to find out all the similar matrices of an given matrix (e.g. matrix $A$)?
And the cardinality of the set that the similar matrices of $A$ composed?
In general, I think, the answer can be complicated, but I can propose a method of describing a set of all matrices that are similar to $A$. I will consider only complex matrices.
Let's denote by $GL_n$ all nonsingular matrices in $\mathbb{C}^{n \times n}$. This is a group and it acts on $\mathbb{C}^{n \times n}$ in following fashion: $G: A \rightarrow G^{-1}AG$ for $G \in GL_n$. All we want is to compute the orbit of $A$ due to this group action, i.e. the set $Orb(A) = \{G^{-1}AG: G \in GL_n\}$. Let's denote $H_A \subset GL_n$ - a subgroup of all nonsigular matrices that commute with $A$ (you can check that it is a subgroup). Then you have a canonical bijection: $GL_n/H_A \rightarrow Orb(A)$ that maps an element $H_A G\in GL_n/H_A$ to $G^{-1}AG$ (quotient means the set of right congruence classes).
Therefore, we need only to compute $H_A$. If $A = \lambda I$ then $H_A = GL_n$ and $Orb(A) \sim GL_n / H_A$ and consists only of one element. If, for example, your matrix has Jordan's form which is diagonal with distinct elements (eigenvalues) then $A$ commutes only with diagonal matrices and therefore $Orb(A) = GL_n / Diag_n$ where $Diag_n$ is a group of diagonal matrices. When your matrix A has diagonal Jordan's form then group $H_A$ consists of matrices that are blockwise diagonal in the basis of Jordan's form (each block corresponds to an eigenvalue). More complicated situation appears when you have non diagonal Jordan's form (maybe I will write later about this).
Edit: If you are asking about the cardinality of $Orb(A)$ then it is possible to say: if $A$ has at least two distinct eigenvalues $Orb(A)$ is continuum. All that is left - to describe the set of all matrices that commute with a Jordan's block.