The nuclear force is isoscalar (charge independent), it does not depend on the 3rd component of the isospin of the nucleon:
$$ I_3 = \frac 1 2 (n_u - n_d) $$
where $n_u$ and $n_d$ are the number of up and down quarks (and a complementary formula for antiquarks). Of course, quarks weren't known in 1935, but because $m_p \approx m_n$, isospin was known to be an approximate symmetry.
The nucleon is then viewed as a single particle with to states:
$$ |I=\frac 1 2, I_3 = \pm \frac 1 2 \rangle $$
corresponding to the proton and neutron.
The pions then form an isospin triplet ($I=1$):
$$ \pi^+ = |u\bar d\rangle \rightarrow |1, +1\rangle $$
$$ \pi^0 = \frac 1{\sqrt 2}(|u\bar u\rangle - |d\bar d\rangle) \rightarrow |1, 0\rangle $$
$$ \pi^- = |d\bar u\rangle \rightarrow |1, -1\rangle $$
so the neutral pion is absolutely necessary to mediate $pp$ and $nn$ coupling. The terms in the nucleon-nucleon potential look like:
$$ V_{a,b} = V + W \vec I_a \cdot \vec I_b $$
which is isoscalar. The $\vec I$ are vectors in isospin space, and have math just like quantum angular momentum, hence the name: isospin.