What is the conserved quantity of a scale-invariant universe?

The symmetry you are asking about is usually called a scale transformation or dilation and it, along with Poincare transformations and conformal transformations is part of the group of conformal isometries of Minkowski space. In a large class of theories one can construct an "improved" energy-momentum tensor $\theta^{\mu \nu}$ such that the Noether current corresponding to scale transformations is given by $s^\mu=x_\nu \theta^{\mu \nu}$. The spatial integral of the time component of $s^\mu$ is the conserved charge. Clearly $\partial_\mu s^\mu = \theta^\mu_\mu$ so the conservation of $s^\mu$ is equivalent to the vanishing of the trace of the energy-momentum tensor. It should be noted that most quantum field theories are not invariant under scale and conformal transformations. Those that are are called conformal field theories and they have been studied in great detail in connection with phase transitions (where the theory becomes scale invariant at the transition point), string theory (the two-dimensional theory on the string world-sheet is a CFT) and some parts of mathematics (the study of Vertex Operator Algebras is the study of a particular kind of CFT).

thank you for the nice question. It directly relates to the topics of conformal field theories. I found a very nice thread in another forum where I guess your question has been answered.
Nevertheless, I will try to summarize the main points here and maybe add some points.

Symmetries in General Relativity

In general relativities, symmetries correspond to an isometry of the metric $g=g_{ab}dx^a dx^b$, say $\varphi^\star g = g$. That means, if you move along the path of such a symmetry, it does not change. This can be expressed in terms of the Lie-derivative.
$$L_v g = 0$$ or $$\nabla_{(a}v_{b)} = 0$$

where the parenthesis stands for symmetrization over indices and $v = \dot\varphi(t)$ is the vector field associated to $\varphi$. One can find very nice introductory calculations for this in Robert M. Wald: General Relativity and Hans Stephani's Introduction to Special and General Relativity.

If $n$ is a unit geodesic, further integration of

$$Q=v_a n^a$$

leads to conserved quantities since $$n^a \nabla_a \left(Q = n^b v_b\right) = n^a n^b\nabla_b v_a + v_b n^a \nabla_a n^b\equiv 0 $$ due to the Killing and geodesic equations.

Famous examples are mass $M$ (or energy) for a stationary spacetime or angular momentum $J$ for axial symmetry (yes, one can assign a spacetime an angular momentum, I found it puzzling in the first place), $$M = 2\int_\Sigma \left( T_{ab}-\frac12 T^n_{\,n}g_{ab} \right)n^a \xi^b dV$$ $$J = -\int_\Sigma T_{ab}n^a \eta^b dV$$ where now $\xi$ is the stationary Killing vector, often $\xi = \partial_t$ and $\eta$, often $\eta=\partial_\varphi$ holds for the axial symmetry and $n$ is now vector perpendicular to a space-like hypersurface $\Sigma$.

Conformal isometries

Now, the situation is a little bit different. A conformal Killing vector $c$ now gives rise to a symmetry of the form $$L_cg=\omega^2g$$ and the conformal Killing equation, implicitly defining $\omega$ now takes the form $$\nabla_{(a}c_{b)} = \frac1n g_{ab}\nabla_d c^d$$

In your case, you force $\omega = 1$ but this is not of great importance as you will see next.

What happens to the "conservation equation"? We have $$n^a \nabla_a \left( n^b c_b\right) = \frac1n \left( \nabla_d c^d \right) n^a n_a$$

which is only zero if $n^a n_a = 0$, a null-geodesic. So, only for a very special class of movements, here light-particles, one will find a symmetry. But this was expected since conformal transformations will not change angles thus light movement won't be affected.

I don't think that this is a conserved quantity in the sense of Emmy Noether.

Sincerely

Robert

PS.: I apologize for any inconvenience concerning notation. I hope everything is clear from context.

Jeff Harvey has of course provided you with the perfect, standardized answer: the scale invariance boils down to the tracelessness of the stress-energy tensor. But the tracelessness is not really a "conserved quantity" in the usual sense that you may have waited for.

However, one may transform the problem to something that is a conserved quantity in the usual sense.

In particular, you may take your scale-invariant universe and insert a point-like object at a chosen point that I will call the origin. In quantum field theory, this is achieved by acting on the vacuum state with a local operator at the origin.

The transformations proving scale invariance are just radial expansions that keep the origin untouched. The laws of physics are invariant under these transformations, by assumption, and this symmetry is equivalent to the conservation of the dimension of the operator from the previous paragraph. But its conservation not with respect to the normal evolution in time but evolution in the "radial time", $\ln(r)$. Consequently, the dimensions of all operators are well-defined in scale-invariant theories. In scale-non-invariant theories, they would depend on the renormalization scale.

I added this verbal exercise in order to emphasize that the scale transformations in a scale-invariant theory are analogous - and in a very well-defined mathematical sense, equivalent - to ordinary translations in time. To be a bit specific, think about 2-dimensional Euclidean theories. The complex coordinate $z$ may be written as $\exp(a+ib)$. Here, $b$ is a periodic, angular variable with periodicity $2\pi$. However, $a$ is real and goes from $-\infty$ to $+\infty$.

The scale transformations are nothing else than the ordinary translations in $a$ which are linked to a Hamiltonian. For example, you expand $z$ $e$-times by shifting $a$ by one. And indeed, scale invariance in 2 dimensions implies the full conformal invariance - under all transformations that preserve the angles - so instead of looking at the $z=x+iy$ plane, you may equally well look at the $a+ib$ plane where the original scale transformation looks like an ordinary translation in the $a$ direction. By conformal symmetry, the form of the action in the $z$ and $a+ib$ coordinates are identical.

In higher dimensions, it is not quite true that scale invariance (and Lorentz/rotational symmetry) implies the full conformal symmetry, but in the important cases, it is true, anyway.

Best wishes Lubos