Let $B\overset{\circ}{=}\left(B_{t}\right)_{t\geq0}$ denote a Brownian motion in a filtration $\mathcal{F}$. Are $X_{t}=\frac{1}{\sqrt{a}}B_{at}$ ($a>0$ constant) and/or $Y_{t}=tB_{\frac{1}{t}}$ Brownian motions in $\mathcal{F}$?
According to my knowledge, the first one is called Scaling-Invaraince property and the second one is called Time-Inversion property. I know they are Brownian motions, but I don't know in which filtration. I would say they are Brownian motions in there own filtrations, so $X$ is Brownian motion in $\mathcal{F}_{X}\overset{\circ}{=}\left(\sigma\left(X_{t}\right)\right)_{t}$ and $Y$ is Brownian motion in $\mathcal{F}_{Y}\overset{\circ}{=}\left(\sigma\left(Y_{t}\right)\right)_{t}$, but not necessary in $\mathcal{F}$.
For example if $\mathcal{F}\overset{\circ}{=}\left(\sigma\left(B_{t}\right)\right)_{t}$, then $B$ is indeed Brownian motion in $\mathcal{F}$, but $X$ can't be. It can't be, because if $a>1$, then at time $t$ we should know the value $B_{at}$, but we “can't see the future”. $B_{at}$ is not adapted to $\mathcal{F}$ and neither $\frac{1}{\sqrt{a}}B_{at}$. The similar holds if we examine $Y$. For example at $t=\frac{1}{2}$ we don't know (nor can calculate) the value of $\frac{1}{2}B_{2}$, therefore it is not adapted to $\mathcal{F}$. To be Brownian motions in $\mathcal{F}$ they should be adapted to $\mathcal{F}$. Is this train of thoughts correct? If it is not, then where is the mistake?
I understand $X$ and $Y$ are Brownian motions, but in which filtration?
$X_t$ and $Y_t$ will be Brownian motions under their natural filtrations; i.e. under $$\mathcal{F}_t^{X} = \sigma(X_s\, : \, s \leq t) \qquad \text{ and } \qquad \mathcal{F}_t^{Y} = \sigma(Y_s\, : \, s \leq t)$$ respectively. As you noted, $X$ and $Y$ may fail to be adapted to $\mathcal{F}_t$, which means they are not $\mathcal{F}_t$-Brownian motions.