Let $A$ be a commutative Artinian ring and suppose that $f \colon A \to A$ is a ring-morphism. Observe that $f(A) \supseteq f^2(A) \supseteq f^3(A) \supseteq \dots$
Is it true that there exists $n_0 \in \mathbb{N}$ such that $f^{n_0}(A) = f^n(A)$ for every $n \geq n_0$?
I would say that the answer is no. I have the following counterexample, but I am not sure of its correctness. Let $F$ be a field and $A \colon\!= F(\{x_n : n\in\mathbb{N}\})$ be the set of rational functions (i.e. formal quotients of polynomials) over the set of countable many variables $\{x_n : n\in\mathbb{N}\}$.
In this context, $A$ is Artinian (in fact, a field) and the map $T\colon \{x_k : n\in\mathbb{N}\} \to A$ defined by $x_k \to x_{k+1}$ extends to a ring-morphism $T\colon A \to A$ (not sure about these two things). Then, $T^n(A) = F(\{x_k : k\geq n\})$, so the sequence of images $T^n(A)$ never stabilizes.
Maybe I should also mention that when $A$ is an Artinian module and $f$ a module endomorphism, then my question has an affirmative answer.
Your counterexample is correct.
That fields are Artinian is clear as there are only two ideals in any fields and fields of rational functions are fields nonetheless. You could argue that the map you propose is induced by universal properties if that makes you feel safer regarding their existence. This can be done by combining the universal property of the polynomial ring in countable many variables and the universal property of the quotient field. This should do the job of formally extending the shifting map to a ring endomorphism as desired.
The reason why this fails is that the image of an ideal along a ring homomorphism is not necessarily an ideal. Hence non of the ring-theoretic properties apply to the chain $A\supseteq f(A)\supseteq f^2(A)\supseteq\cdots$ of subsets. On the other hand, it does work in the case of Artinian modules because the image of a submodule along a homomorphism of modules is again a submodule.