Why symmetry transformations change either the state or the operator but not both. Why? Why is it assumed that the symmetry transformations change either the state $\psi$ or the operator $O$ but not both simultaneously? For example, if you assume $\psi\to T\psi$, you take $O\to O$ and if you assume $\psi\to \psi$, you take $O\to T^{-1}OT$. What is the logic here? This bit is not explained in the textbooks I am using (Griffiths and Shankar).
 A: You have exactly the same situation with time evolution. In Schrödinger picture of quantum mechanics, the state of the system is assumed to evolve according to
$$i\hbar{d\over dt}|\psi(t)\rangle=\hat H|\psi(t)\rangle
\ \Leftrightarrow\ |\psi(t)\rangle=e^{-i\hat Ht/\hbar}|\psi(0)\rangle$$
while operators $\hat O$ are time-independent. Averages evolve according to
$$\langle O(t)\rangle=\langle\psi(t)|\hat O|\psi(t)\rangle
=\langle\psi(0)|e^{i\hat Ht/\hbar}\hat Oe^{-i\hat Ht/\hbar}|\psi(0)\rangle$$
You see that you could obtain the same expression of $\langle O(t)\rangle$ by assuming that the state is time-independent, i.e. $|\psi(t)\rangle=|\psi(0)\rangle$, but the operators evolve according to
$$\hat O(t)=e^{i\hat Ht/\hbar}\hat Oe^{-i\hat Ht/\hbar}
\ \Leftrightarrow\ i\hbar{d\over dt}\hat O(t)=[\hat H,\hat O(t)]$$
We have recovered Heisenberg picture of QM.
Time evolution is a special case of a transformation that can be applied to a quantum state. The general case is obtained by replacing $e^{-i\hat Ht/\hbar}$ by $\hat T$. Either you consider that the state is affected by the transformation as $|{\psi'}\rangle=\hat T|\psi\rangle$ and you assume that operators are not or you consider that the state is invariant but then operators should transform as $\hat O'=\hat T^+\hat O\hat T$ in such a way that the expression of averages
$$\langle O'\rangle=\langle\psi'|\hat O|\psi'\rangle
   =\langle\psi|\hat O'|\psi\rangle
   =\langle\psi|\hat T^+\hat O\hat T|\psi\rangle$$
is recovered. Note that the average $\langle O'\rangle$ after transformation is not equal to $\langle O\rangle$. We just assumed that the two pictures give the same answer.
