A classical result is that solutions of the isoperimetric problem on the plane, the sphere, and the hyperbolic plane are circles. Are there any other Riemannian metrics on these spaces that share this property or does this characterize metrics of constant curvature ?
I should add that by circles I mean really standard circles, not geodesic circles for the new metric.
It is known (and follows from an easy calculation) that solutions of the isoperimetric problem on a surface have constant geodesic curvature. In his 1887 classic Leçons Sur La Théorie Générale Des Surfaces Et Les Applications Géométriques Du Calcul Infinitésimal, G. Darboux states, without indicating a proof, that a surface for which all of the curves of constant geodesic curvature are closed must have constant Gaussian curvature. According to Blaschke, a proof of this was published in 1921, by B. Baule in Math. Ann. 83 (286-310) and 84 (202-215), but I have never checked it out myself. I believe that this answers your question, at least for Riemannian metrics.
Concerning the Finsler case: (I had a little time over the holiday to think about this, and, after a brief calculation, came up with the following answer, which is interesting if you are willing to widen the question to considering 'Finsler metric-measure' spaces.)
Consider the following question: For what choices of Finsler metric $F$ on the plane and area form (i.e., 'measure) $\Omega$ can the solutions of the isoperimetric problem be the standard circles in the $xy$-plane? (Here, the isoperimetric problem is to be understood as finding domains of a given area with minimal perimeter where the 'perimeter' is measured by integrating $F$ over the boundary of the domain and the 'area' is measured by integrating $\Omega$ over the domain.)
The answer turns out to be this: Let $a$, $b$, $c$, $e$, $p$ and $q\not=0$ be constants, and let $D$ be the domain in the $xy$-plane where $a(x^2+y^2) + bx + cy + e >0$. Define the Finsler function $$ F = \phi + \frac{\sqrt{dx^2+dy^2}}{a(x^2+y^2) + bx + cy + e} $$ where $\phi$ is any $1$-form that satisfies $$ d\phi = \frac{p\ dx\wedge dy}{\bigl(a(x^2+y^2) + bx + cy + e\bigr)^2} $$ and has sufficiently small norm that $F$ is positive on nonzero tangent vectors in $D$. Meanwhile, take $$ \Omega = \frac{q\ dx\wedge dy}{\bigl(a(x^2+y^2) + bx + cy + e\bigr)^2} $$
Then the isoperimetric curves in $D$ for the metric-measure structure $(F,\Omega)$ are the standard circles in $D$. Moreover, any metric-measure structure on a domain in the plane that has this property is of the above form.
Note that the case $\phi=0$ is exactly the metrics of constant curvature that are conformal to the standard metric on the plane, as we already knew. The more general case is that of a so-called Randers metric of a particular kind, one whose 'associated' Riemannian metric is also a solution.
It is rather a comment and a question to the author than the answer (but it is too long for the comment window).
What is the (solutions of the) isoperimetric problem? If it is a closed curve of a fixed length such that it bounds the region of a maximal volume, or a shortest closed curve that bounds the region of certain fixed volume, I believe that the solutions of isoperimetric problem are very rare.
In order to confirm my suggestion, I would like to mention that for metrics of constant positive curvature the length of the circle bounding the ball of volume 1 is less than that of for metrics of constant negative curvature. This suggests that for the sphere of revolution such that the maximums of the curvature are the poles the isoperimentric circles (of sufficiently small length) are circles (= boundary of the balls) centered at the poles. Then, in a certain coordinate system they are usual circles.
(To see that the solutions of the isoperimetric problem are rare you could also combine two statements from the answer of Robert Bryant: curves of constant geodesic curvature are usually not closed and isoperimetric curves must have constant geodesic curvature.)
