Given a (smooth, orientable) n-dimensional manifold $M$ with two (pseudo-)Riemannian metrics $g_{1}$ and $g_{2}$ of the same signature that induce the same Levi-Civita connection and satisfy $\sqrt{|\det{g_1}|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n} = \sqrt{|\det{g_2}|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n}$,
is $g_{1}=g_{2}$?
The standard Minkowski metric, transformed by scaling one variable by 1/2, and another by 2, has the usual volume form, and the same Levi--Civita connection (which depends only on the affine structure).
Take two inner products on $\mathbb{R}^n$ given by symmetric matrices $(g_{ij})_{1\leq i,j\leq n}$, $(h_{ij})_{1\leq i,j\leq n}$ such that $\det g_{ij} =\det h_{ij}=1$. They have the same volume forms and Levi-Civita connections though they are different if the matrices are different.
Consider a Riemannian manifold admitting a parallel (with respect to the Levi-Civita connection $\nabla^g$) symmetric bilinear form $\beta$ which is not a multiple of the metric $g$. Then, for an open set of constants $(a,b)\in\mathbb R^2$ $$g_{a,b}:=a g+b\beta$$ is a Riemannian metric, i.e., positive definite. The Levi-Civita connection of $g_{a,b}$ is $\nabla^g$. Moreover, the volume form of $g_{a,b}$ is a constant multiple (depending on $a,b$) of the volume form of $g$, as the volume forms $vol_{a,b}$ are parallel with respect to $\nabla^g$. Clearly, at $(a,b)=(1,0)$ we have $g_{a,b}=g$ and $$\frac{\partial vol_{a,b}}{\partial a}\neq0.$$ Hence, you always find a curve of Riemannian metrics with the same Levi-Civita connection and the same volume form under the above condition. Of course, explicit examples are provided by the answers of Ben and Liviu, but also by product metrics. Analogous arguments apply to the pseudo-Riemann case.
By taking the difference between two different metrics with the same Levi-Civita connection and the same volume form one immediately sees that the space of parallel symmetric bilinear forms must be at least two-dimensional.
