It seems there is no connected complete Riemannian manifold $(M,g)$ which is not flat and such that there is an isometry $f:(M,g)\to(M,\lambda^2g)$, where $\lambda\neq0,1$. I will assume $\lambda>1$, if not take the inverse function.
If there were such an isometry $f$, then for any $p\in M$ the sequence $f^np$ would be Cauchy in the Riemannian distance $d_g$, because $d_g(f^n(p),f^{n+1}(p))=\frac{d_g(p,f(p))}{\lambda^n}$.
As $M$ is not flat there are non zero sectional curvatures so we can find a small geodesic triangle $T$ with sum of angles $S\neq\pi$ (we can consider the triangle to be inside a geodesic surface with nonzero curvature, and then apply Gauss-Bonnet). The sum of angles $S_n$ of the triangle $T_n=f^n(T)$ is just $S_n=S$, because $f$ is conformal. However, if $M$ is complete, then the triangles $T_n$ converge to some point $q$. But $g$-sectional curvatures are bounded in a small neighborhood of $q$, and the $g$-areas of the triangles $T_n$ converge to $0$ so by the Gauss-Bonnet theorem, $S_n$ should converge to $\pi$, which doesn't happen because $S_n=S\neq\pi\;\forall n$.
In the non-complete case there are counterexamples: consider a bump function $f:\mathbb{R}^2\to\mathbb{R}$ which is supported in the ball centered at $(1,0)$ and of radius $0.1$. Now consider the function $g:\mathbb{R}^2\setminus\{0\}\to\mathbb{R}$ given by $g(x)=\sum_{n\in\mathbb{Z}}2^nf\left(\frac{x}{2^n}\right)$.
Then the graph of $g$, as a Riemannian submanifold of $\mathbb{R}^3$, is not flat, but it is invariant by the isometry $x\mapsto2x$ of $\mathbb{R}^3$.