quantum mechanics - Applying a NOT gate on an entangled qubit of equal amplitudes of $|1⟩$ and $|0⟩$

I am currently listening to lecture 11 (Superdense coding) of Michael Nielsen's Quantum Computation for the determined series.

It starts off with 2 qubits, both in the $|0\rangle$ state. Then the "control qubit" is phase shifted by $\pi/4$ through a Hadamard gate. And they are eventually made entangled by the application of c-NOT gate. Now as the control qubit was in equal probability of appearing in either 0 or 1, so (as far as I understand) the target qubit also attains the same phase (i.e. $\pi/4$) as the control qubit (please correct me if I am wrong, though).

However, this only ensures that upon measurement, although each one of them has an equal probability of appearing either in the $|1\rangle$ state or in the $|0\rangle$ state, if one of them emerges to be $|0\rangle$, the other will emerge to $|1\rangle$ state.

However I am confused with application of NOT gate to the control qubit. It was already in the state $$\frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right).$$ What does the application of NOT gate do? Nothing. Its state remains the same! As its state does not change so also the state of its entangled partner does not change.

I understand that somehow they plan on changing both the quits using only one qubit (as they are entangled & changing one in some way will affect change in the other), but I cannot understand how. Please clarify this question.