Isospin for Antiparticles In Quarks & Leptons: An Introductory Course in Modern Particle Physics by Halzen and Martin page 42 reads:

The construction of antiparticle isospin multiplets requires care. It is well illustrated by a simple example. Consider a particular isospin transformation of the nucleon doublet, a rotation through $\pi$ about the 2-axis. We obtain
$$\begin{pmatrix} p' \\ n' \end{pmatrix} = e^{-i \pi (\tau_2 /2)} \begin{pmatrix} p \\ n \end{pmatrix} = -i \tau_2 \begin{pmatrix} p \\ n \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1& 0  \end{pmatrix} \begin{pmatrix} p \\ n \end{pmatrix}.$$
We define antinuclear states using the particle-antiparticle conjugation operator C,
$$Cp=\bar{p}, Cn = \bar{n}$$
Applying $C$ therefore gives
$$ \begin{pmatrix} \bar{p}' \\ \bar{n}' \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1& 0  \end{pmatrix} \begin{pmatrix} \bar{p} \\ \bar{n} \end{pmatrix}.$$
However, we want the antiparticle doublet to transform in exactly the same way as the particle doublet, so that we can combine particle and antiparticle states using the same Clebsch-Gordan coefficients, and so on. We must therefore make two changes...

I do not understand what the issue is, however. What do they mean by "we want... to transform in exactly the same way"?
Didn't they just show that they do transform in exactly the same way?
 A: 
I do not understand what the issue is, however. What do they mean by "we want... to transform in exactly the same way"?
Didn't they just show that they do transform in exactly the same way?

Yes and no; they really didn't! The antibaryon spinors you are looking at are upended! They certainly do not transform well under $T_3$.
Remember, the Gell-Mann Nishijima formula $Q=T_3+ B/2$ rules supreme here! It impels the highest-charged member of the doublet to be upstairs, in the upper position of the two-spinor,
$$  \begin{pmatrix} \bar{n} \\ \bar{p} \end{pmatrix}.$$
But now you have a problem: the rotation matrix you just computed is wrong by a minus sign. If you look at how the specific components transform, you see that
$$ \begin{pmatrix} \bar{n}' \\ \bar{p}' \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ -1& 0  \end{pmatrix} \begin{pmatrix} \bar{n} \\ \bar{p} \end{pmatrix}, $$
disastrously. But, as they explained, you need to be handling baryons and antibaryons even-handedly, isospinwise, as though they were spins.
But all is not lost. If you, in addition, flip the sign of the upper component, you do get the correct transformation law you had before,
$$ \begin{pmatrix}- \bar{n}' \\ \bar{p}' \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1& 0  \end{pmatrix} \begin{pmatrix} -\bar{n} \\ \bar{p} \end{pmatrix},\tag{2.41}$$
and a rephasing (the - sign) will not affect the GM-N formula in any way. Note the action of $T_3$ is now conventional, transformation in "exactly the same way", unlike the ugly monster you started with.
So now you may swing these two spinors, of baryons and antibaryons,  as though they were spin 1/2s, and, e.g. compose two of them like (2.42).
You see then that the antisymmetric product of the spin composition, the singlet, has a counter-intuitive + sign for this symmetric isospin composition; and the antisymmetric isotriplet has a - sign, the $\pi^0$, unlike the symmetric spin triplet: a virtually ritual traditional point of puzzlement to novices.
This is all in a sly attempt to avoid formal mathematical language and the systematic definition of the conjugate representation of su(2).  All the rigmarole that puzzled you is just a low-rent  outline of the conjugate representation, geared to instant-gratification correct formulas. (NB. Remember it well if/when you learn about fermion masses in the SM via coupling to the Higgs field, which makes masses possible for both upstairs and downstairs fermions.)
You might choose the more mainstream path of the conjugate representation, like most intros to the SM, like Li& Cheng, a clearly superior reference, in comparison.  Check that an  equivalent way of appreciating (2.42) is to simply transpose your last equation, (2.40), to
$$ \begin{pmatrix} \bar{p}'  & \bar{n}' \end{pmatrix}=\begin{pmatrix} \bar{p} & \bar{n} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1& 0  \end{pmatrix} .$$
