What does the "scaling invariant" Serrin condition mean? What does it mean that the function space $L^q_tL^p_x$ is invariant under the (3D) Navier-Stokes scaling $u(x,t) \mapsto \lambda u(\lambda x, \lambda^2 t)$ if $2/q + 3/p = 1$?
Does it mean that you compute the integral  $$\left(\int_0^t\left( \left(\int_{\mathbb R^3} (\lambda u(\lambda x, \lambda^2 t))^pdx\right)^{1/p}\right)^qdt\right)^{1/q}$$
and find that it is equal to the one with $\lambda = 1$ only for $2/q + 3/p = 1$?
 A: Yes.

(Body of answer must be 30 characters, and I only entered four. So here are a bit more.)

The Computation:
On $\mathbb{R}^3$, set $y = \lambda x$ and so $dx = \lambda^{-3} dy$. Set $s = \lambda^2 t$ so that $dt = \lambda^{-2} ds$. Plug into you integral expression you have
$$ \left(\int_0^t\left( \left(\int_{\mathbb R^3} (\lambda u(\lambda x, \lambda^2 t))^pdx\right)^{1/p}\right)^qdt\right)^{1/q} = \lambda \left(\int_0^t\left( \left(\int_{\mathbb R^3} (u(y, s))^p\lambda^{-3}dy\right)^{1/p}\right)^q\lambda^{-2} ds\right)^{1/q}$$
Now pull out all the factors of $\lambda$ you get a total of $\lambda^{1 - 3/p - 2/q}$. For scaling invariance the index must equal 0.

In terms of "why" the change $u \mapsto \lambda u(\lambda x, \lambda^2 t)=:v$: if you have that $u$ is a solution to the Navier Stokes equation (with appropriate pressure), then the rescaled function $v$ can be checked to be also a solution to Navier Stokes, with the pressure changing by $p \mapsto \lambda^2 p(\lambda x, \lambda^2 t)$. This is under the assumption there is no external force, which would break scaling.
