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Yes, it is true when assuming that both $H$ and $A$ have pure point spectrum, the case of $A$ with continuous part is a bit more complicated. (However I suspect that $A$ is both self-adjoint and unitary from your comment, so $A^2=I$ implies $\sigma(A) \subset \{\pm 1\}$, however it does not matter below.)

Consider an eigenspace $\cal{H}_E$ of $H$ with eigenvalue $\cal E$. It is invariant under $A$, so $A|_{\cal{H}_E}: \cal{H}_E \to \cal{H}_E$ is still selfadjoint and can be written into a diagonal form with respect to a basis of eigenvectors therein. If it admits $n>1$ eigenvectors in $\cal{H}_E $ (i.e. $\cal E$ is a degenerate energy level), it is easy to construct a self-adjoint operator $B : \cal{H}_E \to \cal{H}_E$ which is not diagonal with respect to the found basis of eigenvectors of $A|_{\cal H_E}$ and satisfies $B^2=I$, so that it does not commute with $A|_{\cal{H}_E}$ and every nontrivial function $f(A)$ of it. The self-adjoint operator $A'= P^\perp + PBP$, where $P$ is the orthogonal projector onto $\cal{H}_E$, is another self-adjoint operator commuting with $H$ which is not a function of $A$ and such that $A'^2=I$, thus $H$ would admit another symmetry against the hypothesis. Thus $\cal{H}_E$ has dimension $1$: no different eigenstates of $H$ with the same eigenvalue exist.

ADDENDUM. Regarding commutativity of $A'$ and $H$, if $P_e$ denotes the orthogonal projector onto the eigenspace of $H$ with eigenvalue $e$, we have $$A' H = (PBP+ P^\perp) H = \left(P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} P_e\right) \sum_{e'} e' P_{e'}$$ $$=P_{\cal E} B P_{\cal E} \sum_{e'} e' P_{e'} + \sum_{e \neq {\cal E}} P_e\sum_{e} e' P_{e'}$$ $$ ={\cal E} P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} P_e$$ $$= \left( \sum_{e'} e' P_{e'}\right) P_{\cal E} B P_{\cal E} + \left(\sum_{e} e' P_{e'}\right)\sum_{e \neq {\cal E}} P_e $$ $$= HA'\:.$$

Yes, it is true when assuming that both $H$ and $A$ have pure point spectrum, the case of $A$ with continuous part is a bit more complicated. (However I suspect that $A$ is both self-adjoint and unitary from your comment, so $A^2=I$ implies $\sigma(A) \subset \{\pm 1\}$, however it does not matter below.)

Consider an eigenspace $\cal{H}_E$ of $H$ with eigenvalue $\cal E$. It is invariant under $A$, so $A|_{\cal{H}_E}: \cal{H}_E \to \cal{H}_E$ is still selfadjoint and can be written into a diagonal form with respect to a basis of eigenvectors therein. If it admits $n>1$ eigenvectors in $\cal{H}_E $ (i.e. $\cal E$ is a degenerate energy level), it is easy to construct a self-adjoint operator $B : \cal{H}_E \to \cal{H}_E$ which is not diagonal with respect to the found basis of eigenvectors of $A|_{\cal H_E}$ and satisfies $B^2=I$, so that it does not commute with $A|_{\cal{H}_E}$ and every nontrivial function $f(A)$ of it. The self-adjoint operator $A'= P^\perp + PBP$, where $P$ is the orthogonal projector onto $\cal{H}_E$, is another self-adjoint operator commuting with $H$ which is not a function of $A$ and such that $A'^2=I$, thus $H$ would admit another symmetry against the hypothesis. Thus $\cal{H}_E$ has dimension $1$: no different eigenstates of $H$ with the same eigenvalue exist.

ADDENDUM. Regarding commutativity of $A'$ and $H$, if $P_e$ denotes the orthogonal projector onto the eigenspace of $H$ with eigenvalue $e$, we have $$A' H = (PBP+ P^\perp) H = \left(P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} P_e\right) \sum_{e'} e' P_{e'}$$ $$=P_{\cal E} B P_{\cal E} \sum_{e'} e' P_{e'} + \sum_{e \neq {\cal E}} P_e\sum_{e} e' P_{e'}$$ $$ ={\cal E} P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} eP_e$$ $$= \left( \sum_{e'} e' P_{e'}\right) P_{\cal E} B P_{\cal E} + \left(\sum_{e} e' P_{e'}\right)\sum_{e \neq {\cal E}} P_e $$ $$= HA'\:.$$

Yes, it is true when assuming that both $H$ and $A$ have pure point spectrum, the case of $A$ with continuous part is a bit more complicated. (However I suspect that $A$ is both self-adjoint and unitary from your comment, so $A^2=I$ implies $\sigma(A) \subset \{\pm 1\}$, however it does not matter below.)

Consider an eigenspace $\cal{H}_E$ of $H$ with eigenvalue $\cal E$. It is invariant under $A$, so $A|_{\cal{H}_E}: \cal{H}_E \to \cal{H}_E$ is still selfadjoint and can be written into a diagonal form with respect to a basis of eigenvectors therein. If it admits $n>1$ eigenvectors in $\cal{H}_E $ (i.e. $\cal E$ is a degenerate energy level), it is easy to construct a self-adjoint operator $B : \cal{H}_E \to \cal{H}_E$ which is not diagonal with respect to the found basis of eigenvectors of $A|_{\cal H_E}$ and satisfies $B^2=I$, so that it does not commute with $A|_{\cal{H}_E}$ and every nontrivial function $f(A)$ of it. The self-adjoint operator $A'= P^\perp + PBP$, where $P$ is the orthogonal projector onto $\cal{H}_E$, is another self-adjoint operator commuting with $H$ which is not a function of $A$ and such that $A'^2=I$, thus $H$ would admit another symmetry against the hypothesis. Thus $\cal{H}_E$ has dimension $1$: no different eigenstates of $H$ with the same eigenvalue exist.

Yes, it is true when assuming that both $H$ and $A$ have pure point spectrum, the case of $A$ with continuous part is a bit more complicated. (However I suspect that $A$ is both self-adjoint and unitary from your comment, so $A^2=I$ implies $\sigma(A) \subset \{\pm 1\}$, however it does not matter below.)

Consider an eigenspace $\cal{H}_E$ of $H$ with eigenvalue $\cal E$. It is invariant under $A$, so $A|_{\cal{H}_E}: \cal{H}_E \to \cal{H}_E$ is still selfadjoint and can be written into a diagonal form with respect to a basis of eigenvectors therein. If it admits $n>1$ eigenvectors in $\cal{H}_E $ (i.e. $\cal E$ is a degenerate energy level), it is easy to construct a self-adjoint operator $B : \cal{H}_E \to \cal{H}_E$ which is not diagonal with respect to the found basis of eigenvectors of $A|_{\cal H_E}$ and satisfies $B^2=I$, so that it does not commute with $A|_{\cal{H}_E}$ and every nontrivial function $f(A)$ of it. The self-adjoint operator $A'= P^\perp + PBP$, where $P$ is the orthogonal projector onto $\cal{H}_E$, is another self-adjoint operator commuting with $H$ which is not a function of $A$ and such that $A'^2=I$, thus $H$ would admit another symmetry against the hypothesis. Thus $\cal{H}_E$ has dimension $1$: no different eigenstates of $H$ with the same eigenvalue exist.

ADDENDUM. Regarding commutativity of $A'$ and $H$, if $P_e$ denotes the orthogonal projector onto the eigenspace of $H$ with eigenvalue $e$, we have $$A' H = (PBP+ P^\perp) H = \left(P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} P_e\right) \sum_{e'} e' P_{e'}$$ $$=P_{\cal E} B P_{\cal E} \sum_{e'} e' P_{e'} + \sum_{e \neq {\cal E}} P_e\sum_{e} e' P_{e'}$$ $$ ={\cal E} P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} P_e$$ $$= \left( \sum_{e'} e' P_{e'}\right) P_{\cal E} B P_{\cal E} + \left(\sum_{e} e' P_{e'}\right)\sum_{e \neq {\cal E}} P_e $$ $$= HA'\:.$$

Yes, it is true when assuming that both $H$ and $A$ have pure point spectrum, the case of $A$ with continuous part is a bit more complicated. (However I suspect that $A$ is both self-adjoint and unitary from your comment, so $A^2=I$ implies $\sigma(A) \subset \{\pm 1\}$, however it does not matter below.)

Consider an eigenspace $\cal{H}_E$ of $H$ with eigenvalue $\cal E$. It is invariant under $A$, so $A|_{\cal{H}_E}: \cal{H}_E \to \cal{H}_E$ is still selfadjoint and can be written into a diagonal form with respect to a basis of eigenvectors therein. If it admits $n>1$ eigenvectors in $\cal{H}_E $ (i.e. $\cal E$ is a degenerate energy level), it is easy to construct a self-adjoint operator $B : \cal{H}_E \to \cal{H}_E$ which is not diagonal with respect to the found basis of eigenvectors of $A|_{\cal H_E}$ and satisfies $B^2=I$, so that it does not commute with $A|_{\cal{H}_E}$ and every nontrivial function $f(A)$ of it. The self-adjoint operator $A'= P^\perp + BP$, where $P$ is the orthogonal projector onto $\cal{H}_E$, is another self-adjoint operator commuting with $H$ which is not a function of $A$ and such that $A'^2=I$, thus $H$ would admit another symmetry against the hypothesis. Thus $\cal{H}_E$ has dimension $1$: no different eigenstates of $H$ with the same eigenvalue exist.

Yes, it is true when assuming that both $H$ and $A$ have pure point spectrum, the case of $A$ with continuous part is a bit more complicated. (However I suspect that $A$ is both self-adjoint and unitary from your comment, so $A^2=I$ implies $\sigma(A) \subset \{\pm 1\}$, however it does not matter below.)

Consider an eigenspace $\cal{H}_E$ of $H$ with eigenvalue $\cal E$. It is invariant under $A$, so $A|_{\cal{H}_E}: \cal{H}_E \to \cal{H}_E$ is still selfadjoint and can be written into a diagonal form with respect to a basis of eigenvectors therein. If it admits $n>1$ eigenvectors in $\cal{H}_E $ (i.e. $\cal E$ is a degenerate energy level), it is easy to construct a self-adjoint operator $B : \cal{H}_E \to \cal{H}_E$ which is not diagonal with respect to the found basis of eigenvectors of $A|_{\cal H_E}$ and satisfies $B^2=I$, so that it does not commute with $A|_{\cal{H}_E}$ and every nontrivial function $f(A)$ of it. The self-adjoint operator $A'= P^\perp + PBP$, where $P$ is the orthogonal projector onto $\cal{H}_E$, is another self-adjoint operator commuting with $H$ which is not a function of $A$ and such that $A'^2=I$, thus $H$ would admit another symmetry against the hypothesis. Thus $\cal{H}_E$ has dimension $1$: no different eigenstates of $H$ with the same eigenvalue exist.