We define a maximally symmetric space as a Riemannian manifold $(M,g)$ that around each point has the maximum number of $n(n+1)/2$ linearly independent infinitesimal isometries, $n=\dim M$. Equivalently, there exist $n(n+1)/2$ linearly independent Killing vector fields. For such a space one can show that the Riemann curvature tensor can be written as
$$R_{ijkl} = \frac{S}{n(n-1)}(g_{ik}g_{jl} - g_{il}g_{jk}),$$
where $S$ is the scalar curvature, which must be constant.
Now a lot of (mainly physics) references tell me that this formula implies that two maximally symmetric Riemannian manifolds having the same scalar curvature $S$ must be locally isometric. I haven't been able to find a single proof of the statement, however, even after consulting maybe dozens of references.
So assuming the statement is true and does not tacitly need more additional assumptions, how would one prove this? References would be most appreciated as well.
Once you know that both manifolds have the same constant sectional curvature, it follows by using the Jacobi equation to calculate the form of the metric in geodesic normal coordinates. You can find a proof in my book Introduction to Riemannian Manifolds (2nd ed.), Corollary 10.15.