Scaling and time inversion for Brownian motion basically the same? Let $B(t)$ be a Brownian motion. For $a>0$, we have the scaling relation
$$\hat{B}(t)=aB(t/a^2) \sim B(t)$$
and $\hat{B}(t)$ is also a Brownian motion.
The time inversion formula states that
$$B^*(t):= tB(1/t)\mathbb I_{\{t>0\}}, \qquad B^*(0):=0$$
is again a Brownian motion.
Now, if one consideres the formulas pointwise in $t$, then for $a=t$ the statements are identical.
What exactly is the difference in the two properties?
 A: The identities in distribution you recall are of little interest for each given $t$ and one could imagine a host of others: for example, each $B(t)$ is equal in distribution to $\bar B(t)=t^2B(1/t^3)$. 
But the two results you recall are much stronger in that they state that the processes $B=(B(t))_t$, $\hat B=(\hat B(t))_t$ and $B^*=(B^*(t))_t$ have the same distribution. Thus, for every $n\geqslant1$ and $t_1\lt t_2\lt\cdots\lt t_n$, the three random vectors $(B(t_k))_{1\leqslant k\leqslant n}$, $(\hat B(t_k))_{1\leqslant k\leqslant n}$ and $(B^*(t_k))_{1\leqslant k\leqslant n}$ have the same distribution.
In the example we began with, $B(2)-B(1)$ is centered normal with variance $1$ while $\bar B(2)-\bar B(1)=4B(1/8)-B(1)=3B(1/8)-(B(1)-B(1/8))$ is centered normal with variance $9(1/8)+(7/8)=2$, hence the processes $B$ and $\bar B$  have different distributions.
A: Note:
$c$ is a fixed constant, it cannot vary with time.
The scaling property tell you how Brownian motion scales. There are a family of such stochastic processes known as strictly stable processes, which scales to a different factor rather than 1/2. 
It basically says if you shrink the time scale and stretch the process by an appropraite factor, it is still brownian motion.
Time inversion - by its name is 'inverting' time, while scale by $t$
A: If you look at your formula pointwise in $t$,  you look at a much weaker statement, whereas your results are equalities in law for processes. Actually, pointwise, $B_t$ is equal in law to any gaussian with variance $t$.
But your statement is valid also for the processes, which is a stronger result.
To better understand the difference between scaling and inversion of time, the classical intuitive explanation is to imagine that you draw a brownian path with a pen. The scaling is equivalent to zoom or dezoom. The inversion of time, as the name suggests, corresponds simply to draw from the right to the left. And you should get basically the same picture.
