It seems there is no connected complete Riemannian manifold $(M,g)$ which is not flat and such that there is an isometry $f:(M,g)\to(M,\lambda g)$, where $\lambda>0,\lambda\neq1$. 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 $S_n=S\neq\pi\;\forall n$.
For the non complete case it would be enough if there is a point $p$ such that $f^n(p)$ is convergent to some $q$ (that implies that $f^n(x)\to q$ for all $x$, so we can do the same as in the complete case).