I am confused as to how to approach this. So I know the answer is it is isomorphism for a surjective homomorphism but not for an injective homomorphism.
For surjective homomorphism, I used the fact that $A \cong A/\ker f$. Since $f$ is surjective, I think that $\ker f = 0$ but why exactly I am not sure. Also I have not used the Noetherian condition so far. So if I start from assuming $\ker f$ as an ideal, we know that $\ker f \subset I$ where I is some ideal of ring $A$ because of the Noetherian condition.
As for the injective mapping $f$, we can definitely say that $f(r_1)=f(r_2)\implies r_1=r_2$. Therefore, there can not be more than one element in $\ker f$ so this should have been an isomorphism but the test marked this as wrong which means my conclusion that $\ker f = 0$ is incorrect but I can't see why. I don't even know where the Noetherian condition plays a role in this.