If any two triangles of equal area can be mapped via affine maps, what can we say about the geometry? This is a cross-post.
Let $(M,g)$ be a two-dimensional compact surface, endowed with a Riemannian metric.
Fix $s>0$, and suppose that for any two geodesic triangles $A,B$ having area $s$, there exists an affine onto map $f:A \to B$, where I say $f$ is affine if $\nabla df=0$. (equivalently, $f$ maps parametrized geodesics to parametrized geodesics.
Here $\nabla=\nabla^{T^*M} \otimes \nabla^{f^*TM}$).
I assume $s<<\text{Area}(M)$ is very small, so there are a lot of triangles of area $s$.
What can we say about the metric $g$? Does it have to be flat? Are there any restrictions on its curvature?
I do not require $f$ to be the restriction of an affine map $M \to M$;
(I think this is a stronger requirement than the existence of "local" or piece-wise affine maps. e.g. for the flat torus, globally we only have $SL_2(\mathbb{Z})$.)
I believe that the assumption means that that there a lot of affine maps locally $M \to M$; perhaps we can translate this into showing $M$ is flat.
If $\nabla^{T^*M} \otimes \nabla^{f^*TM}$ has zero curvature, then $M$ is flat. And 'many affine maps' means roughly 'many parallel sections of $T^*M \otimes TM$ '-- although not exactly, since for every $f$, $df$ is a section of a vector bundle which depends on $f$, i.e. $T^*M \otimes f^*TM$.
 A: Using the structure equations, it is not difficult to show that, if $f:(M,g)\to(N,h)$ is a diffeomorphism of (not necessarily complete) connected surfaces that is affine in the OP's sense, i.e., $\nabla(\mathrm{d}f)=0$, then $f$ has constant singular values and $L\bigl(f(p)\bigr) = K(p)/|\det(f)|$ for any $p\in M$, where $K:M\to\mathbb{R}$ and $L:N\to\mathbb{R}$ are the Gauss curvatures of $g$ and $h$ respectively, and $\det(f)$ is the product of the singular values of $f$ (and thus is constant).
Moreover, if we restrict attention to the open sets $M^*$ and $N^* = f(M^*)$ where the respective Gauss curvatures are nonvanishing, then $f:M^*\to N^*$ is a homothety, i.e., an isometry up to a (constant) scale factor.
Thus, the only situation in which the OP's desired flexibility holds is for a flat surface.
Here is a little bit more detail:  It's a local calculation, so choose a $g$-orthonormal coframing $g={\omega_1}^2+{\omega_2}^2$ on $U\subset M$ and an $h$-orthonormal coframing $h={\eta_1}^2+{\eta_2}^2$ on $V = f(U)\subset N$.  Let $\omega_{12}$ on $U$ and $\eta_{12}$ on $V$ satisfy
$$
\mathrm{d}\omega_1=-\omega_{12}\wedge\omega_2\quad
\mathrm{d}\omega_2=+\omega_{12}\wedge\omega_1\quad
\mathrm{d}\omega_{12}=-L\,\omega_1\wedge\omega_2
$$
and
$$
\mathrm{d}\eta_1=-\eta_{12}\wedge\eta_2\quad
\mathrm{d}\eta_2=+\eta_{12}\wedge\eta_1\quad
\mathrm{d}\eta_{12}=-K\,\eta_1\wedge\eta_2\,.
$$
There will exist functions $a_{ij}$ on $U$ such that
$f^*(\eta_i) = a_{ij}\,\omega_j$.  The condition $\nabla(\mathrm{d}f) = 0$ then translates into the equations
$$
\mathrm{d}
\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}
= \begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}
\begin{pmatrix}0&\omega_{12}\\-\omega_{12}&0\end{pmatrix}
-\begin{pmatrix}0&\bar\eta_{12}\\-\bar\eta_{12}&0\end{pmatrix}
\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}
$$
where $\bar\eta_{12} = f^*\eta_{12}$.  This immediatly implies the equations
$$
\mathrm{d}(a_{11}a_{22}-a_{12}a_{21}) = 
\mathrm{d}(a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2) = 0,
$$
so the singular values of the matrix $a$ are constant.
If the singular values are equal, it follows that $f$ is either the constant map or else $f$ is a homothety.  If they are not equal, we can choose our coframings so that $0<a_{11}<a_{22}$ and $a_{12}=a_{21}=0$.  Now, the constancy of the $a_{ij}$ together with the above equations implies that we must have
$$
a_{11}\omega_{12}-a_{22}\bar\eta_{12} = a_{22}\omega_{12}-a_{11}\bar\eta_{12} = 0.
$$
In other words $\omega_{12} \equiv \bar\eta_{12} \equiv 0$.  Now the above structure equations imply that $K \equiv L \equiv 0$.
