If $X$ and $Y$ are finite etale covers of each other, are they isomorphic? This is a somewhat strange question perhaps, but I was wondering the following:
Suppose $X$ and $Y$ are schemes (lets say Noetherian, and integral as well if you'd like). Suppose that there are maps $f: Y \rightarrow X, g: X \rightarrow Y$ which realize $Y$ as a finite etale cover of $X$, and vice versa. Does it follow that $X$ and $Y$ have to be isomorphic?
I would be happy with any result along these lines. That is, if it's not always true, are there nice situations where it is? Knowing about the case where $X, Y$ are assumed affine would be just fine too.
 A: No. For instance, there are non-isomorphic elliptic curves that satisfy this relationship (which is called "isogeny" in that case). I'm not sure of any non-drastic simplifying assumptions that make this true.
A: This is just a comment, but I don't have enough reputation yet. 
It's true for compact hyperbolic curves (=smooth projective curves of genus at least two) for trivial reasons. In fact, if $X\to Y$ is a finite etale morphism of compact hyperbolic curves of degree $d$ and $Y\to X$ is a finite etale morphism of degree $e$ we obtain that $$2g_X -2 = (2g_Y-2)d = (2g_X-2)dd^\prime.$$ This implies that $dd^\prime =1$ and thus $d=d^\prime = 1$.
More generally, the above approach shows the following. Suppose that $f:X\to Y$ is finite etale and $g:Y\to X$ is finite etale, where $X$ and $Y$ are smooth projective varieties. Then, the "Riemann-Hurwitz formula" implies that $$K_X = f^\ast K_Y = f^\ast g^\ast K_X = (gf)^\ast K_X.$$ Maybe this implies that $K_X$ is torsion, but I'm not sure. Of course, I'm assuming $\deg g, \deg f>1$.
It might be more natural to just consider the isomorphism $(gf)^\ast \Omega^1_X \to \Omega^1_X$. You can compute the Euler characteristic of $X$ and $Y$ by taking the $n$-th Chern class ($n=\dim X=\dim Y$), and then you see that the Euler characteristic of $X$ (and $Y$) is zero (if the above surjective map is an isomorphism). 
