When A and B are similar matrices，what conditions hold A = C.B When A and B are similar matrices，In which cases，A = C.B

I'm not sure whether this is a valid mathematical question...
$A=D^{-1}B D=C B$
 A: Sure. It makes sense. So your question is: given $B$ any square matrix and $D$ invertible, when does there exist $C$ such that $D^{-1}BD=CB$? Note that all these matrices must be of the same size $n\times n$ for the question to make sense.

Trivial partial answer: if $B$ is invertible, $C$ exists and must be the commutator $D^{-1}BDB^{-1}$.

Now if $B$ is not invertible, the question is whether the linear map $D^{-1}BD$ in $L(K^n)$ associated with this matrix factors through $B$.

Fact: given $S:E\longrightarrow F$, and $T:E\longrightarrow G$ linear, there exists $U:G\longrightarrow F$ linear such that $S=UT$ if and only if $\ker T\subseteq \ker S$.

Proof: the only if is trivial. For the if, just note that if $\ker T\subseteq \ker S$, then $UTx:=Sx$ defines without ambiguity a linear map $U$ on $\mbox{im } T$. Then extend $U$ by  $0$ on an algebraic complement of $\mbox{im } T$ in $G$. This $U$ works. $\Box$.
Therefore, in your case, such a $C$ exists if and only if 
$$\ker B\subseteq \ker D^{-1}BD=D^{-1} (\ker B).$$
Since you are in finite dimension and since both subspaces have the same dimension: 

Such a $C$ exists if and only if
  $$
\ker B=D^{-1}(\ker B)\quad\iff\quad \ker B=D(\ker B).
$$
  There are two cases where this is automatically verified: $\ker B=\{0\}$ i.e. $B$ invertible, which we have seen earlier; and $\ker B=K^n$, i.e. $B=0$, which is not fascinating. If $\ker B$ is nontrivial, this may or may not happen. The condition means that $D$ leaves the nullspace $\ker B$ invariant, that's all. If you are more interested in the matrix $A=D^{-1}BD$, the condition is simply
  $$
\ker A=\ker B
$$
  and it would just be $\ker B\subseteq \ker A$ if $A$ was not assumed to be similar to $B$, condition which forces their nullspaces to be isomorphic.

