Let $A/\mathbb F_q$ be an abelian variety over a finite field with Weil numbers $q^{1/2}\alpha_1,\dots,q^{1/2}\alpha_n$.
Consider the numbers $q^{d/2}\alpha_1,\dots,q^{d/2}\alpha_n$. These are still conjugate Weil numbers and by Honda-Serre-Tate should correspond to an abelian variety $\tilde A/\mathbb F_{q^d}$. What is the relation between $\tilde A$ and $A$?
Consider the $nd$ numbers $\{q^{1/2}\alpha_i^{1/d}\}$. The Abelian variety $\tilde A'$ corresponding to these numbers is, I believe the Weil Restriction of $\tilde A$ from $\mathbb F_{q^d}$ to $\mathbb F_q$. Is there a direct relation between $\tilde A'$ and $A$?
In my particular case, I also have a $\mathbb Z[\zeta_{\ell^m}]$ action on $A$ which has dimension $\phi(\ell^m)$ for $\ell $ a a prime different from $q$, $m$ very large, $d = \ell$ and $\dim \tilde A = \ell\dim \tilde A'$. I don't know if this is relevant at all.
(Of course, Honda-Tate is only upto isogeny but I would be very happy if there was a functor upto isomorphism that picked out some canonical $\tilde A,\tilde A'$.)
Edit - An explanation of the terminology:
Weil numbers are algebraic integers that have absolute value $q^{w/2}$ under every complex embedding where $q$ is a prime power. The integer $w$ is denoted it's weight. They are interesting because the Riemann hypothesis shows that over a finite field $\mathbb F_q$, for any smooth variety $X$, the eigenvalues of the Frobenius on the etale cohomology $H^i(X,\mathbb Q_\ell)$ are Weil numbers of weight $i$.
In particular, since the cohomology of an Abelian variety is determined by $H^1$, the weight $1$ Weil numbers determine all the others.
Conversely, a theorem by Honda-(Serre-)Tate shows that any conjugacy class (under the Galois action) of Weil numbers of absolute value $q^{1/2}$ can be realized in the cohomology of an Abelian variety over $\mathbb F_q$.
