Symmetry transformations: a doubt about the relations that we assume true When we deal with symmetry transformations in quantum mechanics we assume true that, 
If before the symmetry transformation we have this 
$ \hat A | \phi_n \rangle  = a_n|\phi_n \rangle,$
and after the symmetry transformation we have this 
$ \hat A' | \phi_n' \rangle  = a_n'|\phi_n' \rangle,$
then $a_n'=a_n$. 
I think the reason for this relation is that $\hat A$ and $\hat A'$ are equivalent observables (for example the energy in two different frame of references).
The problem is that, if $\hat A=\hat X$ where $\hat X$ is the position operator, then this relation seems wrong, because we would have:
$ \hat X | x \rangle  = x|x \rangle$ and
$ \hat X' | x' \rangle  =x|x' \rangle$ both ture, 
that means that the position eigenstate seen by two different frames of references is seen in the same coordinates. How can this be true if the systems are for example translated one to the other?
 A: *

*This is not an assumption, it is a requirement for consistency. The symmetry transformation acts on operators and states, it does not act on numbers. So the equation $A\lvert \psi_n \rangle = a_n\lvert \psi_n\rangle$ simply becomes $A'\lvert \psi_n'\rangle = a_n\lvert \psi_n'\rangle$ after applying the transformation. This equation must be true for any linear transformation on the space of states, regardless of whether it is a symmetry or not. 

*So when the transformation is a translation by $a$, it acts as $\hat{x}\mapsto \hat{x} - a$ on the position operator and $\lvert x\rangle \mapsto \lvert x + a\rangle$ on its eigenstates. The equation $\hat{x}\lvert x\rangle = x\lvert x\rangle$ becomes $(\hat{x}-a)\lvert x + a\rangle = x\lvert x + a\rangle$. There is nothing inconsistent about this - note that the transformed equation does not claim that $\lvert x + a\rangle$ would be a position eigenstate with eigenvalue $x$, but instead says that $\lvert x + a\rangle$ is an eigenstate of $\hat{x}-a$ with eigenvalue $x$.
A: A symmetry transformation leaves the scalar products invariant, therefore
$$
a_n = \langle \phi_n| A |\phi_n\rangle = \langle \phi'_n|A'|\phi'_n\rangle = a_n'\,.
$$
It is not a contradiction with the position operator because you are measuring a transformed position in a transformed frame. Let me put it this way: suppose $x$ is the origin, so when you measure it with $X$ you get zero. Then make a translation to a frame where the origin is $x'=\vec{a}$. If you measure the position $\vec{a}$ in the new frame, you still get zero, because that's the origin in that frame.
A: Adding a bit to the answer of @MannyC using explicitly the translation operator:
If $T(a)$ is such that $T(a)\vert x\rangle = \vert x'\rangle = \vert x+a\rangle$, then $T(a) \hat x T^{-1}(a)=\hat x'=\hat x-a\hat 1$ so that
\begin{align}
x'=\langle\hat x'\vert\ \hat x'\vert  x'\rangle &=
\langle x\vert T^{-1}(a) T(a) \hat x T^{-1}(a) T(a)\vert x\rangle 
=\langle x\vert \hat x \vert x\rangle = x\, , \\
&=\langle x+a\vert (\hat x-a\hat 1)\vert x+a\rangle=
\langle x+1\vert  x+a\rangle\left(x+a -a\right)=x
\end{align}
What this tells you is that the average value of the translated position operator in the translated basis is the same as the average value of the original position operator in the original basis.
A: To get rid of the confusion it can be beneficial to think about QM in terms of Linear Algebra. Your operators are just matrices and your states vectors. 
If you act with a matrix on it's eigenvector, you get out the corresponding eigenvalue to the eigenvector. If you transform your matrix and eigenvector into a different basis, you will still get the same eigenvalue, as the eigenvalues are more general than the representation of the matrix and vector in a certain basis. 
