Does the isometry group determine the Riemannian metric? Suppose $G \subset \text{Iso}(M)$ is a Lie group acting smoothly on a (pseudo-)Riemannian manifold $(M, g)$. Then $G$ induces a Lie algebra of Killing vectors on $M$. In this paper by Goenner and Stachel (in particular in appendix B), they mention how to find a coordinate representation for the metric out of the Killing vectors and the Casimir invariant. However, I do not understand what they mean by "solving the structure equations for the Killing vector fields". Certainly one cannot find unique solutions to the structure equations since adding multiples of commuting vector fields yields a new solution. The system of first order PDE's that one obtains from the coordinate representations of the Lie brackets are unwieldy at best, so it seems one needs to incorporate the Killing equation in order to work through the problem. Yet, the Killing equation depends on the metric, which is precisely what we are trying to find. What exactly are the authors trying to say?
 A: I think that you are missing some hypotheses on the action of $G$, otherwise there are trivial counterexamples.  For example, if $G\subset\mathrm{Iso}(M,g)$ is the trivial group, then one clearly cannot reconstruct $g$ from just knowing $G$ as a subgroup of $\mathrm{Diff}(M)$.  Beyond that, the most you can hope for is to construct $g$ up to constant multiples, since $\mathrm{Iso}(M,cg) = \mathrm{Iso}(M,g)$ for any nonzero constant $c$.
A more reasonable question is to ask how to tell whether, given a (connected) Lie subgroup $G\subset\mathrm{Diff}(M)$ or, equivalently, its Lie algebra ${\frak{g}}\subset\mathrm{Vect}(M)$, one can tell whether or not there is a (pseudo-Riemannian) metric $g$ such that $G\subset\mathrm{Iso}(M,g)$ and, if so, provide an algorithm for finding $g$.
Some hint that this might be tricky can be seen by considering the torus $M = \mathbb{R}^2/\mathbb{Z}^2$ and letting the group be $G\simeq\mathbb{R}$ acting as  $t\cdot[x,y] = [x {+} t, y {+} at]$ for some irrational constant $a$.  The only $G$-invariant metrics are of the form $g = E\,\mathrm{d}x^2 + 2F\,\mathrm{d}x\,\mathrm{d}y+G\,\mathrm{d}y^2$ where $E$, $F$, and $G$ are constants with $EG-F^2\not=0$.  On the other hand, if $a$ were rational, then $G$ would be isomorphic to $S^1 = \mathbb{R}/\mathbb{Z}$ and there would be arbitrary functions worth of  $G$-invariant metrics.
In general, you don't expect $G$ to determine a unique metric up to multiples unless $G$ acts transitively on $M$, and, even then, the metric may not be unique up to constant multiples.  A good example, is $M = G$ where $G$ acts on itself by left translations.  Then any left-invariant metric on $G$ will do (and each of them is completely determined by its value at the identity of $G$).
Meanwhile, in the transitive case, it's not hard to determine whether there is an invariant metric and whether or not it is unique.  You just have to look at the (closed) subgroup $G_p\subset G$ consisting of the elements of $G$ that fix some given $p\in M$.  That group acts linearly on $T_pM$, and you just need to know whether that linear representation of $G_p$ fixes a non-degenerate inner product on $T_pM$.  If $G_p$ is connected, this is a purely linear algebra problem, since one can easily compute the Lie algebra of $G_p$ as represented in $T_pM$ and then, by solving a linear system of equations, find the subspace $S_p$ of $G_p$-invariant quadratic forms on $T_pM$.  Then, assuming that this subspace contains a non-degenerate quadratic form, you are done.  If $G_p$ is not connected, you first find the corresponding subspace $S^\circ_p$ for $G_p^\circ$, the identity component of $G_p$, and then deal with the problem of finding the $G_p/G_p^\circ$-invariant elements of $S_p^\circ$, again, an algebra problem.
