Preface
One commonly finds this interaction Lagrangian in phenomenology (ignoring constants):
$$\mathscr{L} \propto \bar\psi \gamma^5\gamma^\mu \vec\tau\psi \cdot \partial_\mu \vec\pi \tag{1}$$
Evaluating the scalar product yields
$$\vec\tau \cdot \vec \pi = \begin{pmatrix}\pi_3 & \pi_1 -i \pi_2\\ \pi_1 + i \pi_2 & -\pi_3 \end{pmatrix} = \begin{pmatrix}\pi_0 & \sqrt{2}\pi^+ \\ \sqrt{2}\pi^- & -\pi_0 \end{pmatrix} \tag{2}$$
Here we've used the eigenstates of $I_z$, the generator for rotations around the z-axis, i.e. $\pi^\pm = \frac{1}{\sqrt 2}(\pi_1 \mp i \pi_2)$, $\pi_0 = \pi_3$
This introduces a factor of $\sqrt{2}$ at the vertices with charged pions and a factor of $-1$ for the neutron-neutron interaction. I am curious where these factors come from, because I can not reproduce it with Clebsch-Gordon coefficients.
Naive Attempt
Supposing I want to write down an interaction for a $0^-$ scalar particle with a derivative coupling, I could do it as such
$$\mathscr{L} \propto \bar\psi \gamma^5\gamma^\mu \psi \partial_\mu \phi$$
Now without knowing about the first Lagrangian, how would I introduce isospin? For $\phi = \pi$ and a nucleon, I can think of the following interactions:
$$
\begin{align}
\pi^0 p &\rightarrow p \tag{4a}\\
\pi^- p &\rightarrow n \tag{5a}\\
\pi^0 n &\rightarrow n \tag{6a}\\
\pi^+ n &\rightarrow p \tag{7a}
\end{align}
$$
So naively I could write down for Lagrangians
$$
\begin{align}
\mathscr{L} &\propto \bar\psi_p \gamma^5\gamma^\mu \psi_p \partial_\mu \pi^0\tag{4b}\\
\mathscr{L} &\propto \bar\psi_n \gamma^5\gamma^\mu \psi_p \partial_\mu \pi^-\tag{5b}\\
\mathscr{L} &\propto \bar\psi_n \gamma^5\gamma^\mu \psi_n \partial_\mu \pi^0\tag{6b}\\
\mathscr{L} &\propto \bar\psi_p \gamma^5\gamma^\mu \psi_n \partial_\mu \pi^+\tag{7b}
\end{align}
$$
but now the isospin factors are missing. How can I derive them? Just "guessing" $\tau_a \pi_a$ seems a bit contrived.
My attempt
Writing this down in braket notation, I would find the interaction $\pi^-p\rightarrow n$ to be something like
$$
\begin{align}
\langle f| O |i \rangle &= \left\langle \tfrac{1}{2}, -\tfrac{1}{2}\right| O \left|1, -1; \tfrac{1}{2}, \tfrac{1}{2}\right\rangle \tag{8}\\
&= \sum \limits_{M=-\tfrac{1}{2}}^{\tfrac{1}{2}} \left\langle \tfrac{1}{2}, -\tfrac{1}{2}\right| O \left|\tfrac{1}{2},M\right\rangle\underbrace{\left\langle\tfrac{1}{2},M\,\right|\left.1, -1; \tfrac{1}{2}, \tfrac{1}{2}\right\rangle}_{=C^*}\tag{9}\\
&= -\sqrt{\tfrac{2}{3}}\left\langle \tfrac{1}{2}, -\tfrac{1}{2}\right| O \left|\tfrac{1}{2}, -\tfrac{1}{2}\right\rangle\tag{10}
\end{align}
$$
Here in $(9)$ we have added a 1 as a sum over the compound state, which gives us the clebsch-gordan coefficient $-\sqrt{\tfrac{2}{3}}$.
But I do not really know how to continue.
Best Answer
You are looking down on the 2nd column of the Clebsch matrix, up to an over-all common normalization of $-1/\sqrt 3$, ignored here.
Rewrite (1),(2) as the sum of (4b,5b,6b,7b), $$ \bar\psi \gamma^5\gamma^\mu \vec\tau\psi \cdot \partial_\mu \vec\pi \\ = \bar\psi_p \gamma^5\gamma^\mu \psi_p \partial_\mu \pi^0 + \sqrt{2} \bar\psi_n \gamma^5\gamma^\mu \psi_p \partial_\mu \pi^- - \bar\psi_n \gamma^5\gamma^\mu \psi_n \partial_\mu \pi^0 + \sqrt{2} \bar\psi_p \gamma^5\gamma^\mu \psi_n \partial_\mu \pi^+ , $$ And you are trying to connect the relative coefficients with this second column of the Clebsch matrix, namely the projection onto isospin 1/2 of the $1\otimes 1/2$, nucleon-pion state; the respective coefficients of the terms in this order being $$ \langle \tfrac{1}{2}, \tfrac{1}{2} | 1, 0 \tfrac{1}{2}, \tfrac{1}{2} \rangle = -1/\sqrt{3}; \\ \langle \tfrac{1}{2}, -\tfrac{1}{2} | 1, -1; \tfrac{1}{2}, \tfrac{1}{2} \rangle =-\sqrt{2/3}; \\ \langle \tfrac{1}{2}, -\tfrac{1}{2} | 1, 0; \tfrac{1}{2}, -\tfrac{1}{2} \rangle =1/\sqrt{3} ; \\ \langle \tfrac{1}{2}, \tfrac{1}{2} | 1, 1; \tfrac{1}{2}, -\tfrac{1}{2} \rangle = -\sqrt{2/3}. $$ The last Clebsch may need rephasing, but you see that the relative coefficients comport with the top expression you see in books. This is where the relative strengths and phases of (4a,5a,6a,7a) come from.