This is a perfect storm of notational dissonance between QM and QFT. Your statement
I had read that the symmetry is spontaneously broken if $A \left|\psi \right>_n^{(1)}\neq 0 $ and symmetric if $A \left|\psi \right>_n^{(1)}= 0. $
is inapposite and misconstrued—justly paradoxical. I suspect labels on symmetric phase versus SSB might not be productive here.
- Review of SSQM
$$
Q=\begin{pmatrix} 0& 0\\ A& 0 \end{pmatrix}, \qquad Q^\dagger=\begin{pmatrix} 0& A^\dagger\\ 0& 0 \end{pmatrix},
$$
so
$$
H= Q Q^\dagger+ Q^\dagger Q= \begin{pmatrix} A^\dagger A& 0\\ 0& A A^\dagger \end{pmatrix}\equiv \begin{pmatrix} H_1& 0\\ 0& H_2 \end{pmatrix}.
$$
Now
$$
A|\psi_n^{(1)}\rangle = |\psi_{n-1}^{(2)}\rangle, \qquad A^\dagger |\psi_n^{(2)}\rangle = |\psi_{n+1}^{(1)}\rangle, \qquad E^{(1)}_{n+1}= E_n^{(2)} \qquad E_0^{(1)}=0.
$$
So $\psi^{(2)}_n$ and $\psi^{(1)}_{n+1}$ have the same eigenvalue under H, and they are degenerate pairs-- (unbroken symmetry),
$$
Q \begin{pmatrix} \psi^{(1)}_{n+1}\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ \psi^{(2)}_{n} \end{pmatrix} .
$$
But the ground state is unique, an unpaired susy-singlet, and non-degenerate:
$$
Q \begin{pmatrix} \psi^{(1)}_{0}\\ 0 \end{pmatrix} = 0, \qquad A|\psi^{(1)}_0\rangle = 0 ,
$$
so, if you chose, you might call it SSBroken (but Witten calls it unbroken). A Susy rotation does not couple it with other states. So the bottom of the spectrum, the ground state, lacks the symmetry of the rest of the spectrum.
This is a narrow answer to your question, but does not alleviate the perfect storm of dissonance.
- The QFT paradigm The above stands in sharp contrast to QFT with its infinite d.o.f.
To avoid confusion, I'll call the infinitesimal Susy charge here ${\cal Q}$, so that the real correspondent to the finite Q above is, instead, the full group
super-transformation, $\exp i\bar\theta {\cal Q}$, for a Grassmann angle θ. In QFT, the symmetric (unbroken) phase is characterized by
$$
{\cal Q}|0\rangle= 0, \Longrightarrow \exp (i\bar\theta {\cal Q})|0\rangle=|0\rangle .
$$
For $[H,{\cal Q}]=0$, eigenstates of the hamiltonian,
$$
H \phi |0\rangle= E \phi |0\rangle
$$
are susy-rotated to degenerate ones,
$$
H(e^{i\bar\theta {\cal Q}} \phi e^{-i\bar \theta {\cal Q}} )|0\rangle =
e^{i\bar\theta {\cal Q}} H\phi |0\rangle = E (e^{i\bar\theta {\cal Q}} \phi e^{-i\bar \theta {\cal Q}} )|0\rangle,
$$
evocative of "what you had read". So, it is here that this is the hallmark of degeneracy of the spectrum, while the vacuum is unique, similarly to above.
In the SSB phase, we have the opposite: most states are non-degenerate, by failure of the above argument, but the vacuum is now degenerate. It is not unique:
$$
{\cal Q}|0\rangle\neq 0 \Longrightarrow \qquad |\Omega\rangle= \exp (i\bar\theta {\cal Q})|0\rangle\neq |0\rangle .
$$
This state is degenerate with the vacuum and roils and bubbles with goldstinos.
So, at the end of the day, your question dramatizes the stark difference between QM and QFT. The takeaway may well be that sticking to facts rather than labels might be the sanest option.