Let $g_t$ be the solution of normalized Ricci Flow on a Riemann surface $\Sigma$ of genus g. We know that $g_t$ converges to constant curvature metric. Is it possible for $g_t$ to be of the form $f_t^*g_0$, for a one parameter family of diffeomorphisms $f_t$?

2$\begingroup$ Since $f_t^*g_0$ is always isometric to $g_0$, if the limit has constant curvature, so does $f_tg_0$ for all time $t$ and therefore so does $g_0$. In that case, the normalized Ricci flow does nothing, and $f_t$ is the identity map for all $t$. $\endgroup$ – Deane Yang Mar 22 '13 at 15:16
If the Riemann surface is compact, then the Ricci flow converges to a constant curvature metric, but if not compact it might not converge. If you don't suitably rescale to keep the same volume, then the flow will not converge, as the volume will scale to infinity or zero. If you suitably rescale, then the flow will converge and the scalar curvature will satisfy a heat equation, forcing its maximum to be strictly decreasing. Moreover, the convergence is uniform with all derivatives, so the scalar curvature will uniformly converge to a constant. But the curvature of $f_t^* g_0$ has the same maximum and minimum curvature that $g_0$ did, and by sequential compactness, the location of the maximum must converge after taking a subsequence of $t$ values going to infinity, and similarly the minimum, so the curvature is not converging to a constant.
Let $(M,g)$ be a (bad) Riemannian $2$orbifold with one cone point $x_{1}$ of order $p\geq2$ and underlying space a topological $2$sphere, i.e., a teardrop with Euler characteristic $\chi(M)=1+\frac{1}{p}$. For $\varepsilon>0$ small, we have length $\operatorname{L}(\partial B_{\varepsilon}(x_{1}))\approx2\pi p^{1}\varepsilon$ and geodesic curvature $\kappa(\partial B_{\varepsilon }(x_{1}))\approx\varepsilon^{1}$. The GaussBonnet formula is, using $\chi(MB_{\varepsilon}(x_{1}))=1$ and the scalar curvature $R=2K$ where $K$ is the Gauss curvature, $$ \int_{M}Rd\mu=\lim_{\varepsilon\rightarrow0}\int_{MB_{\varepsilon}(x_{1} )}Rd\mu=4\pi+2\lim_{\varepsilon\rightarrow0}\int_{\partial B_{\varepsilon }(x_{1})}\kappa ds=4\pi\chi(M). $$
A vector field $X$ is smooth on $M$ if it is smooth on $Mx_{1}$ and its $p$fold lift via the orbifold chart $\pi:B\rightarrow U$, where $U$ is a neighborhood of $x_{1}$, is smooth (the same applies to functions). This implies that $\lim_{x\rightarrow x_{1}}X\left( x\right) =X\left( x_{1}\right) =0$. Moreover, the divergence theorem is true for $X$: $\int _{M}\operatorname{div}X\,d\mu=0$. In particular, if $u$ is a smooth function on $M$, then $\int_{M}\Delta u\,d\mu=0$.
Conversely, we may solve $\Delta f=h$ if (and only if) $\int_{M}hd\mu=0$. E.g., use the calculus of variations to obtain a minimizer $f\in W^{1,2}(M)$ of the functional $J\left( f\right) =\int_{M}(\frac{1}{2}\left\vert \nabla f\right\vert ^{2}+hf)d\mu$ with the constraint $\int_{M}fd\mu=0$ (so we can apply the Poincare inequality; the constraint anchors the function). By regularity theory, the minimizer $f$ is smooth and satisfies $\Delta f+h=\operatorname{const}$, which must be zero. This works on an orbifold with isolated singularites.
Let $r$ be the average scalar curvature. Since $\int_{M}(rR)d\mu=0$, there exists $f$ (unique up to an additive constant) such that $\Delta f=rR$, called the curvature potential. Note that $\nabla f\left( x_{1}\right) =0$.
The only closed bad $2$orbifolds are the teardrop and the spindle (i.e., a bad orbifold with two cone points, of orders $p,q\geq2$, where $p\neq q$ and $\chi=\frac{1}{p}+\frac{1}{q}$). On these, LangFang Wu proved that for any initial smooth metric with positive curvature, the normalized Ricci flow (NRF) asymptotically aproaches a (nonconstant curvature) shrinking gradient Ricci soliton. More specifically, consider the NRF modified by diffeomorphisms, defined by $\frac{\partial}{\partial t}g=2\left( \operatorname{Ric} +\nabla^{2}f\frac{r}{2}g\right) $, where the curvature potential $f$ satisfies $\frac{\partial f}{\partial t}=\Delta f\left\vert \nabla f\right\vert ^{2}+rf$. (The right side of the first equation is actually tracefree, which implies that the area form is pointwise constant under the flow). Then, as $t\rightarrow\infty$, $g\left( t\right) $ and $f\left( t\right) $ converge to a $C^{\infty}$ metric $g_{\infty}$ and a $C^{\infty}$ function $f_{\infty}$, each exponentially fast in all $C^{k}$norms. They satisfy $\operatorname{Ric}(g_{\infty})+\nabla_{g_{\infty}}^{2}f_{\infty }\frac{r}{2}g_{\infty}=0$ and $\Delta_{g_{\infty}}f_{\infty}\left\vert \nabla f_{\infty}\right\vert _{g_{\infty}}^{2}+rf_{\infty}=0$.
So, Wu proved that under the NRF on a bad closed $2$orbifold there exists a gradient Ricci soliton structure $(\tilde{g}_{\infty},\tilde{f}_{\infty})$ equivalent to $(g_{\infty},f_{\infty})$ such that $g\left( t\right) $ asymptotically approaches $\varphi_{t}^{\ast}\tilde{g}_{\infty}$, where $\{\varphi_{t}\}$ is generated by $\operatorname{grad}_{\tilde{g}_{\infty} }\tilde{f}_{\infty}$. If the initial metric yields a shrinking gradient Ricci soliton structure, then we have equality, i.e., $g\left( t\right) =\varphi_{t}^{\ast}g\left( 0\right) $.
Disclaimer: I haven't been thinking about orbifolds recently, so some of the above is what I think is true from memory without checking the details; please look out for errors.