(nontrivial) isotrivial family of elliptic curves I think it should be a standard procedure to construct such things, can anyone give a reference or give a hint? Can this be done over any base scheme?
 A: (Edit: added motivation for my "answer" even though it isn't exactly what was asked for. It shows the moduli space of genus 1 curves cannot be fine - see last paragraph.) 
The Hopf surface is a (non-algebraic) example of a non-trivial family of isomorphic elliptic curves. Here I use "elliptic curve" to mean "smooth complex curve of genus 1", which is probably not what you mean. (In particular my curves have no distinguished base point.)
The Hopf surface $X$ is a quotient of $\mathbb{C}^2\setminus0$ by the action of $\mathbb Z$ generated by $z \mapsto 2z$. Since this action commutes with the action of $\mathbb{C}^*$ there is a map $X \to \mathbb{CP}^1$. Each fibre is a copy of $\mathbb{C}\setminus\{0\}$ divided by the action $z\sim 2z$. So all the fibres are isomorphic elliptic curves. (If you want base points on each curve, this example doesn't work because the fibration $X \to \mathbb{CP}^1$ doesn't have a section.) On the other hand, its easy to see that $X$ is diffeomorphic to $S^1\times S^3$ and so you can't trivialise the fibration even topologically. Replacing $z\mapsto 2z$ by $z\mapsto \lambda z$ for $\lambda\in \mathbb{C}$ non-zero, I guess you can get any genus-1 curve to appear as all fibres of a topologically non-trivial family. 
Even though this is not eaxctly what you're looking for, I thought it worth giving the example anyway, because it's a very simple way to see that the moduli space of genus 1 curves cannot be fine. (Fine moduli spaces carry a universal family and all other families are pulled back from the universal family via the map to moduli space - in particular a family of isomorphic objects in such a moduli space is pulled back by the constant map and so  must be trivial. One often sees that objects cannot be parametrised by a fine moduli space by giving examples of special objects with "extra" automorphisms. The Hopf surface is an alternative to that approach in this situation.)
A: Hint: use quadratic twists.
Edit: So as not to drag things out, I hope it's okay if I just give you a standard example.  Let
$E_0: y^2 = x^3 + Ax + B$
be your favorite elliptic curve over $\mathbb{Q}$ (i.e., any will do).  Consider the 
elliptic curve
$E: t y^2 = x^3 + Ax + B$
over the rational function field $\mathbb{Q}(t) = \mathbb{Q}(\mathbb{P}^1)$.  Spreading this out as a scheme over $\mathbb{P}^1_{/\mathbb{Q}}$, we see that there are two singular fibers, at $t = 0$ and $t = \infty$.  Discarding these we get an elliptic curve over $\mathbb{A}^1 \setminus \{0\}$ which is isotrivial -- the $j$-invariant over every fiber is $j(E_0)$ -- but nontrivial: the isomorphism classes of the fibers are in bijection with $H^1(\mathbb{Q},\mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Q}^{\times}/\mathbb{Q}^{\times 2}$.
It's a good bet that you'll find this example somewhere in the chapter on elliptic surfaces in Silverman's Advanced Topics in the Arithmetic of Elliptic Curves.
Can it be done over any base scheme?  Unless I misunderstand, of course not, e.g. not over the spectrum of a field.
