Can we extend any flat metric on the torus to a "standard" global flat metric? Let $\mathbb{T}^d=\mathbb{R}^d/\mathbb{Z}^d$ be the $d$-dimensional torus. Let $p \in M$, and let $U \subseteq \mathbb{T}^d$ be an open neighbourhood of $p$ diffeomorphic to $\mathbb{R}^d$. Let $g_U$ be a flat Riemannian metric defined on $U$.

Can $g_U$ be extended to a smooth metric $h$ on $\mathbb{T}^d$ which is isometric to one of the "standard" flat metrics on $\mathbb{T}^d$ induced by the Euclidean metric on $\mathbb{R}^d$? 

In other words, I ask if there exist a metric $h$ on $\mathbb{T}^d$ such that


*

*$h$ coincides with $g_U$ on $U$.

*There exist a global isometry $\phi:(\mathbb{T}^d,h) \to (\mathbb{T}^d,g_0)$ where $g_0$ is one of the standard metrics  which come from the identification $\mathbb{T}^d \simeq \mathbb{R}^d/\mathbb{Z}^d$.


The reason I said "one of" is that there are multiple possible options for how to identify $\mathbb{T}^d \simeq \mathbb{R}^d/\mathbb{Z}^d$-instead of the standard squares one can use parallelograms with non-straight angles. (I think that different angles will give non-isometric metrics on the Torus).
I am fine with shrinking $U$ if that's necessary.
Edit:
An easier question is whether or not all flat metrics on the torus are isometric?
 A: By shrinking $U,$ we can guarantee that $U$ is a smooth $d$-cell (isometrically) embeddable as a bounded region in $\mathbb R^d,$ and thus embeddable in $(\mathbb T^d, \delta)$ where $\delta$ is a large enough standard flat metric on the torus. Letting $\phi : U \to \phi(U)$ denote this isometry, we can extend $\phi$ to a diffeomorphism $\tilde \phi$ of $\mathbb T^d$. The metric $\tilde \phi^* \delta$ is then an extension of $g_U$ that is isometric to the standard metric $\delta.$
As an aside, it is indeed true that any flat metric on a torus is isometric to a standard one: If you have a flat metric $g$ on $\mathbb T^d,$ then this lifts to a flat metric $\tilde g$ on the universal cover, which must thus be isometric to Euclidean $\mathbb R^d.$ The deck transformation group $\pi_1(\mathbb T^d) \simeq \mathbb Z^d$ of this cover acts by $\tilde g$-isometries, so $(\mathbb T^d,g) = (\mathbb R^d,\tilde g)/\mathbb Z^d$ is the quotient of a Euclidean space by an isometric $\mathbb Z^d$-action.
The answer to the "easy question" is that isometry classes of flat metrics on $\mathbb T^d$ are in 1-1 correspondence with isometry classes of lattices in $\mathbb R^d,$ so indeed choosing a differently shaped fundamental domain will give you a non-isometric metric.
