I am trying to understand the relation between Wigner's $3j$-symbols (or Clebsch-Gordon coefficients) and matrix coefficients of intertwiners. I am new to this topic and need some help to understand it properly.

So, let us start with the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$, which is the complexification of $\mathfrak{su}(2)$. Now, as usual, all the irreducible representations of $\mathfrak{sl}(2,\mathbb{C})$ can be described by spins $j\in\mathbb{N}_{0}/2$ and have dimension $2j+1$. Now it is a general fact that for a simple and complex Lie algebra, ever finite-dimensional irreducible representation can be labeled with a heighest weight $\Lambda$, which in the case of $\mathfrak{sl}(2,\mathbb{C})$ are given by $2j$. In general, the tensor product of two such heighest weight modules is fully reducible and hence we can write $$V_{\Lambda}\otimes V_{\Lambda^{\prime}}=\bigoplus_{i}C_{\Lambda\Lambda^{\prime}}^{\Lambda_{i}}V_{\Lambda_{i}}$$ with some coefficients, usually called multiplicities or Littlewood-Richardson coefficients. Now in order to relate this to intertwiners, we first of all know, according to the Lemma of Schur, that the space of intertwiners between two irreducible finite-dimensional representations of a complex Lie algebra $\mathfrak{g}$ is either $0$-dimensional (if they are not isomorphic) or $1$-dimensional (if they are isomorphic). As a consequence, we get that the Littlewood-Richardson coefficients are given by the dimension of the space of intertwiners from $V_{\Lambda_{i}}$ to $V_{\Lambda}\otimes V_{\Lambda^{\prime}}$, i.e.

$$C_{\Lambda\Lambda^{\prime}}^{\Lambda_{i}}=\mathrm{dim}_{\mathbb{C}}(\mathrm{Int}(V_{\Lambda_{i}},V_{\Lambda}\otimes V_{\Lambda^{\prime}}))$$

So in the case of $\mathfrak{sl}(2,\mathbb{C})$ (or equivalently $\mathfrak{su}(2)$), the coefficients are hence given by

$$C_{\Lambda\Lambda^{\prime}}^{\Lambda_{i}}=\mathrm{dim}_{\mathbb{C}}(\mathrm{Int}(V_{2j_{i}},V_{2j}\otimes V_{2j^{\prime}}))$$

So far so good. In a textbook (Fuch, Schweigert - Symmetries, Lie Algebras and Representations), they then say that using this correspondence, it is clear that the Clebsch-Gordon coefficients for fixed $J$ are the matrix coefficients of intertwiners in $\mathrm{Int}(V_{2j_{i}},V_{2j}\otimes V_{2j^{\prime}})$ for "a definite choice of basis". Can anyone explain this step to me in more detail? Furthermore, what does "a definite choice of basis" mean in this context? Also, what is then the correspondence with the $3j$-symbols?

Thank you very much!

EDIT: In order to clarify the conventions I use, let me add some more details about the Clebsch-Gordon coefficients: Let $J_{\pm},J_{0}$ be the three generators of $\mathfrak{sl}(2,\mathbb{C})$, i.e. $$[J_{0},J_{\pm}]=\pm J_{\pm}\hspace{1cm}\text{and}\hspace{1cm}[J_{+},J_{-}]=2J_{0}$$ Then there is a basis $\{\mid j,m\rangle\}_{-j\leq m\leq j}$ of $V_{2j}$ satisfying $$J_{0}\mid j,m\rangle=m\mid j,m\rangle$$ $$J_{\pm}\mid j,m\rangle=\sqrt{j(j+1)-m(m\pm 1)}\mid j,m\pm 1\rangle$$ In order to define the Clebsch-Gordon coefficients, one usually defines to different bases of the tensor product $V_{2j}\otimes V_{2j^{\prime}}$:

(1) The tensor product of the bases described above, i.e. $\{\mid j_{1},j_{2},m_{1},m_{2}\rangle\}_{-j_{1}\leq m_{1}\leq j_{1},-j_{2}\leq m_{2}\leq j_{2}}$ where $\mid j_{1},j_{2},m_{1},m_{2}\rangle:=\mid j_{1},m_{1}\rangle\otimes\mid j_{2},m_{2}\rangle$.

(2) Define the total spin $\vec{J}:=\vec{J}_{1}+\vec{J}_{2}$. Then there is a basis $\{\mid J,M\rangle:=\mid j_{1},j_{2},J,M\rangle\}$ satisfying $$\vec{J}^{2}\mid J,M\rangle=J(J+1)\mid J,M\rangle$$

Then the Clebsch-Gordon coefficients are the coefficients of the change of basis matrix, i.e. $\langle j_{1},j_{2},m_{1},m_{2}\mid J,M\rangle$