Let $k$ be an algebraically closed field. Let $R,S$ be rings such that $R,S$ both contain $k$ as a subring. Let $\varphi:R\to S$ be a ring homomorphism. Then does $\varphi(a)=a$ for all $a\in k$?
So far I think: $\varphi:R\to S$ is a ring homomorphism $\implies \varphi|_{k}:k\to k$ is a field homomorphism $\implies \varphi(1_{k})=1_{k}$.
But I think this only shows that $\varphi$ fixes elements of $k$ generated by $1_{k}$. So let's say $k$ is $\mathbb{C}$ (I don't know any other algebraically closed field). Then I reckon that this would imply $\varphi$ fixes $\mathbb{Z}$ (and by the answer I got from this, $\mathbb{Q}$ as well), but I don't know what I can say about irrational or imaginary numbers.
The case where all the rings are $\mathbb C$ is already interesting enough! As Berci mentioned in the comments, conjugation gives an example of an automorphism that only fixes the reals. One can show that this is the only non-trivial automorphism fixing the reals.
To answer your second question in the comments: Assuming choice, there are even more exotic counterexamples, so called wild automorphisms. You might find the discussion on this MSE post interesting.