How to find a matrix that transforms vectors from basis $B$ into vectors from basis $C$? $$B=\left\{\begin{bmatrix}
1 \\
2 \\
3 
\end{bmatrix},
\begin{bmatrix}
3 \\
2 \\
1 
\end{bmatrix},
\begin{bmatrix}
1 \\
1 \\
0 
\end{bmatrix}\right\} 
\quad C=\left\{\begin{bmatrix}
4 \\
5 \\
6 
\end{bmatrix},
\begin{bmatrix}
2 \\
4 \\
7 
\end{bmatrix},
\begin{bmatrix}
6 \\
7 \\
8 
\end{bmatrix}\right\} $$
I need to find such a matrix 
$A=\begin{bmatrix}
    a & b & c \\
    d & e & f \\
    g & h & i \\
    \end{bmatrix}$ that transforms vectors from $B$ into vectors from $C$, so for example:
$$\begin{bmatrix}
    a & b & c \\
    d & e & f \\
    g & h & i \\
    \end{bmatrix} \begin{bmatrix}
    1 \\
    2 \\
    3 \\
    \end{bmatrix}=\begin{bmatrix}
    4 \\
    5 \\
    6 \\
    \end{bmatrix}\text{ etc...}$$ 
I want to understand the general idea of solving such tasks, so please don't give away the answer!
 A: Here's a good way to start:
let
$$
E = \Bigg\{\begin{bmatrix}
1 \\
0 \\
0 
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
0 
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
1 
\end{bmatrix}\Bigg\} 
$$
and ask "Can I find a transformation taking the $E$ vectors to the $B$ vectors?" The answer is "sure, just put the $B$ vectors into the columns of a 3x3 matrix." 
You should do this and convince yourself that this matrix, which I'll call $M_{E\to B}$does in fact take the $E$ vectors to the corresponding $B$ vectors. 
Now...what matrix would do the opposite, would take the $B$ vectors to the $E$ vectors? 
It's $M_{E\to B}^{-1}$, of course! So we now know that 
$$
M_{B \to E} = M_{E \to B}^{-1}.
$$
Now you can do the same thing and find a matrix that takes the $E$ vectors to the $C$ vectors, right? 
Now what happens if you multiply the $E$ vectors by the matrix
$$
Q = M_{E\to B}M_{C \to E}?
$$
If you take it one step at a time, you'll see that $c_1$ gets send to $e_1$, which then (by the first matrix) gets sent to $b_1$. 
Try this out on a $2 \times 2$ example to convince yourself it works --- it's a lot easier to invert a $2 \times 2$ matrix!
A: Consider the following so called commutative diagram:
$$\require{AMScd}\begin{CD}
\mathbb R^3 @>A>> \mathbb R^3\\
@A(b_1 b_2 b_3)AA @AA(c_1 c_2 c_3)A \\
B @>>I> C \end{CD}$$
If you have a vector with respect to basis $B$, such as (1,0,0), this vector is actually $\mathbf b_1$, the first vector in the basis of $B$. More generally, (x,y,z) with respect to $B$ is $x\mathbf b_1+y\mathbf b_2+z\mathbf b_3$. It means that we can convert a vector with respect to $B$ to the standard basis of $\mathbb R^3$ by simply putting the basis vectors of $B$ as columns in a matrix.
In your case, you're interested in the conversion from $B$ to $C$, which is effectively the identity matrix $I$. That is, for instance (1,0,0) with respect to $B$, which is $\mathbf b_1$, must be mapped to (1,0,0) with respect to $C$, which is $\mathbf c_1$.
To find the corresponding matrix $A$ with respect to the standard basis, we need to follow the arrows, and in reverse where necessary:
$$A = \begin{pmatrix}\mathbf c_1&\mathbf c_2&\mathbf c_3\end{pmatrix}
\cdot I \cdot\begin{pmatrix}\mathbf b_1&\mathbf b_2&\mathbf b_3\end{pmatrix}^{-1}
$$
Reading from right to left, we start with a vector in the standard basis, convert it to $B$ by the corresponding inverse, multiply by $I$, and convert from $C$ to the standard basis.
