If you're talking about a single-qubit transformation where the Bloch vector is changed from $\vec{n}$ to $-\vec{n}$, then I would way think about how the Bloch vector is changing. You want it to go from
$$
(n_x,n_y,n_z)\rightarrow(-n_x,-n_y,-n_z).
$$
However, if we do any unitaries we want on the initial state, this doesn't change the feasibility of the operation, so let me start by applying pauli-$Y$. Hence we have achieved the transformation
$$
(n_x,n_y,n_z)\rightarrow(-n_x,n_y,-n_z),
$$
and the only step left is an effective
$$
(n_x,n_y,n_z)\rightarrow(n_x,-n_y,n_z).
$$
However, *this* is the transpose operation which, given the question, I'm assuming you already know to be impossible, and hopefully know how to prove is impossible (probably based on completely positive maps).

This is a reduction to a known problem. You could probably prove it more directly now. For example, take a suitable two-qubit state and show that the action of $T$ on one qubit does not leave a proper quantum state.

1I'm a bit confused about what exactly the state $|n\rangle$ (and conversely, $|-n\rangle$) exactly is. I assume it's not $|-n\rangle = |(-1)n\rangle = (-1)|n\rangle$? Because then it is just an (irrelevant) global phase... – JSdJ – 2020-11-11T18:59:56.233

1Also, for HW problems, I'd recommend you show some work / attempt to talk about your underlying logic – C. Kang – 2020-11-11T19:18:22.220

Suppose there are N superpositions. Then I guess |-n> denotes |N - n> for n = 0, 1, ..., N-1. – David Leiden – 2020-11-11T20:06:53.247

1@DavidLeiden As far as I can tell, that's not what the operator $T$ does (or if your definition is correct, at least refer to a document that contains this definition -- the one you attach above doesn't). It seems that the action of $T$ is to invert the Bloch vector, namely, take a state $|\vec{n}\rangle \overset{T}{\mapsto} |-\vec{n}\rangle$; as an example, it should flip the $\sigma_z$ eigenstates $|\vec{z+}\rangle \overset{T}{\mapsto} |\vec{z-}\rangle$. For example, $\sigma_x$ would be a state-inversion operator for these states. – keisuke.akira – 2020-11-12T02:11:31.087

However, it wouldn't be an inversion operator for its own eigenstates $|\pm\rangle$ (since its action is trivial, up to a phase) -- and hence it is not universal in its action of flipping. If this is what the $T$ operator means then you need to show that there is no universal flip operator (just as I showed above by

exampleand not proof). – keisuke.akira – 2020-11-12T02:11:44.2171@keisuke.akira after some thought I also think this is the case. Your argument together with imposing linearity on $T$ could already proof the statement though. – JSdJ – 2020-11-12T08:50:20.870

@JSdJ I agree. Namely, if we're allowed to assume that the linear operator $T$ is normal, then it has eigenvectors and its action on them is trivial (up to global phases) and so $T$ is

notuniversal (even though it isphysical). On the other hand, if weassumeuniversality, we can then show that, similar to no-go theorems, this would violate linearity (or unitarity; although, as far as I can tell, the inner product is preserved by this map, at least on the Bloch sphere). – keisuke.akira – 2020-11-12T09:31:08.320