quantum field theory - Detail on C vs. CP violation

In the answer given by knzhou to the post What distinguishes the behaviour of particle from its antiparticle: C violation or CP violation? it is said that

"but the reaction $i \rightarrow f$ will run at the same rate as its $CP$ conjugate $\bar{f}_P \rightarrow \bar{i}_P$."

My questions is: is not $\bar{f}_P \rightarrow \bar{i}_P$ the $CT$ of $i \rightarrow f$ instead of its $CP$ conjugate since you have the anti-$f$ particles in the initial state?

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EDIT I found something else that I do not understand. Why does knzhou say that C is not enough? If C distingishes particles and antiparticles and our theory violates C (not necessarily CP), the reactions $i\rightarrow f$ and its C conjugate $\bar{i} \rightarrow \bar{f}$ will have different rate since $i, f$ and its counterparts $\bar{i}, \bar{f}$ are different, so why do we need CP and not just C?

Moreover, if CPT is preserved but we look for CP violation, does it imply that T is not preserved? But is it not T a symmetry usually?