Matrix of endomorphism in different basis Let $E$ be a vector space, with $\dim E = n$.
Let $u$ be an endomorphism of $E$ and $B = (b_1, \dots, b_n)$, $C = (c_1, \dots, c_n)$ be two basis of $E$.
Let $\operatorname{Mat}_{BC}(u)$ denote the matrix of $(u(b_1), \dots, u(b_n))$ in basis $C$.
Here is my question: if we introduce $B', C'$ two basis of $E$, is this formula right ? 
$$\operatorname{Mat}_{B'C'}(u) = \left(\operatorname{Mat}_{C'C}(\operatorname{Id}_E)\right)^{-1}\operatorname{Mat}_{BC}(u)\operatorname{Mat}_{B'B}(\operatorname{Id}_E)$$
Secondly, if this is correct, any explanation or interpretation would be greatly appreciated.
 A: Yes, your formula is correct although it is usually written differently. Let me introduce some notation which will make the formula look more natural. Given an $n$-dimensional vector space $E$ and an ordered basis $B = (b_1, \dots, b_n)$ for $E$, any vector $v \in E$ can be written uniquely as $v = v_1 b_1 + \dots + v_n b_n$ where $v_i \in \mathbb{F}$ are called the compomenets of the vector $v$ with respect to the basis $B$. Set
$$ [v]_B = \begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix}. $$
That is, $[v]_B$ is a column vector which contains the components of the vector $v$ with respect to the basis $B$. 
Now, let's assume $E,F$ are vector spaces, $B = (b_1, \dots, b_n)$ is a basis for $E$ and $C = (c_1, \dots, c_m)$ is a basis for $F$. Given a linear map $T \colon E \rightarrow F$, there is a unique $m \times n$ matrix denoted by $[T]^B_C$ (you should read it as the matrix representing $T$ with respect to the basis $B$ in the domain and the basis $C$ in the codomain) which satisfies
$$ [Tv]_C = [T]^B_C \cdot [v]_B $$
for all $v \in V$. On the right we have the matrix multiplication of the $m \times n$ matrix $[T]^B_C$ by the column $n \times 1$ vector $[v]_B$ so we get a $m \times 1$ column vector which contains the components of $Tv$ with respect to the basis $C$. The way to remember this formula (which is essentially the definition of $[u]_C^B$) is to note that the $B$'s cancel out as in:
$$ \require{cancel} [Tv]_C = [T]^{\cancel{B}}_C \cdot [v]_{\cancel{B}}. $$
Now, let's say we have another linear map $S \colon F \rightarrow G$ and a basis $D = (d_1, \dots, d_l)$ for $G$. The fact that matrix multiplication corresponds to composition of linear maps can be written as
$$ \require{cancel} [S \circ T]^B_D = [S]_D^{\cancel{C}} \cdot [T]_{\cancel{C}}^B $$
where again, on the right you have matrix multiplication and on the left you have the matrix representing the composition.
In particular, if $E = F$ and $T$ is invertible, then by taking $S = T^{-1}$ we get
$$ \require{cancel}  I_n = [\operatorname{Id}_E]_B^B = [T^{-1} \circ T]^B_B = [T^{-1}]_B^{\cancel{C}} \cdot [T]_{\cancel{C}}^B $$
and so
$$ [T^{-1}]_B^{C} = \left( [T]_{C}^B \right)^{-1}. $$

The relation between the notation I described and your notation is that $\operatorname{Mat}_{BC}(u) = [u]^B_C$. Translating your formula to my notation, we get
$$ [u]^{B'}_{C'} = \left( [\operatorname{Id}_E]^{C'}_{C} \right)^{-1} \cdot [u]^B_C \cdot [\operatorname{Id}_E]^{B'}_B = [\operatorname{Id}_E^{-1} ]^{C}_{C'}  \cdot [u]^B_C \cdot [\operatorname{Id}_E]^{B'}_B = [\operatorname{Id}_E ]^{\cancel{C}}_{C'}  \cdot [u]^{\cancel{B}}_{\cancel{C}} \cdot [\operatorname{Id}_E]^{B'}_{\cancel{B}} $$
so your formula is just an expression of the fact that matrix multiplication corresponds to composition.
