If the matrix rings over two Rings of the same size are isomorphic, then the scalar rings are isomorphic

Question

Let $R$ and $R'$ be rings (with 1 but no further assumptions) and $n \in \mathbb{N}$. Does the following implication hold?

If $M_n(R) \simeq M_n(R')$ then $R \simeq R'$.

If the rings are commutative then it follows from considering the centers of $M_n(R)$ and $M_n(R')$. If the rings are division it also holds and even stronger, it holds even if the matrices are not of the same size.

I thought about considering the embeddings of $R$ in $M_n(R)$ as diagonal matrices, but since I do not want to asume anything about the isomorphism I was not able to conclude that these subrings of matrices in $M_n(R)$ and $M_n(R')$ are isomorphic. I hope someone can help me or with a proof, an idea or a counterexample. Thank you.

Answer 1

There are counterexamples.

I'm not sure how easy the simplest examples are, but you could look at

Chatters, A. W., Non-isomorphic rings with isomorphic matrix rings, Proc. Edinb. Math. Soc., II. Ser. 36, No. 2, 339-348 (1993). ZBL0796.16022.

which has examples (not the first), and whose introduction gives a good overview of previous examples.

Answer 2

According to a comment by Tom Goodwillie in this MO question, there exist rings $R,S$ such that $M_2(R) \cong M_4(S) = M_2(M_2(S))$ yet $R \ncong M_2(S)$.