You wrote the following:
In a boosted frame moving at velocity v, $$L' = \gamma L$$ (from length contraction)
Yes. Know what else is moving with velocity
v? From [POST=2928447]Masterov's problem[/POST], which I was replying to (emphasis added):
I remind to you of a school-task about the two foot-passengers and the dog:
1. Two travelers go on the road with the same velocity ($$v$$) at a distance ($$L$$) from each other (one behind the other).
2. A dog runs between travelers (at velocity $$c$$).
QUESTION: how long time the dog ran ahead, and how long - ago.
That makes
L the contracted length, and
L' the rest length.
In a boosted frame, given frame measurements L, $$L' = L/ \gamma$$
No, you've misunderstood the formula. It relates the rest length of an object to the (contracted) length of the object in a frame in which it is moving. More technically, it relates the distance between two parallel worldlines in a frame in which they're at rest (vertical on a Minkowski diagram) to their spatial distance in some other frame.
You can derive this from the Lorentz transformation.
From Masterov's problem, you can describe the travellers with the trajectories
$$
\begin{eqnarray}
x_{1} &=& vt \,+\, L \,,
x_{2} &=& vt \,.
\end{eqnarray}
$$
A Lorentz boost of velocity
v along the
x axis is:
$$
\begin{eqnarray}
t' &=& \gamma ( t \,-\, \frac{v}{c^{2}} x )
x' &=& \gamma ( x \,-\, v t ) \,.
\end{eqnarray}
$$
You can invert this (if you don't already know the answer) to get the inverse Lorentz boost:
$$
\begin{eqnarray}
t &=& \gamma ( t' \,+\, \frac{v}{c^{2}} x' )
x &=& \gamma ( x' \,+\, v t' ) \,.
\end{eqnarray}
$$
Substituting this into the travellers' trajectories, you get
$$
\begin{eqnarray}
\gamma( x'_{1} \,+\, v t' ) &=& \gamma v ( t' \,+\, \frac{v}{c^{2}} x'_{1} ) \,+\, L \,,
\gamma( x'_{2} \,+\, v t' ) &=& \gamma v ( t' \,+\, \frac{v}{c^{2}} x'_{2} ) \,.
\end{eqnarray}
$$
Dividing both sides by $$\gamma$$ and collecting the
xs on the left side gets you
$$
\begin{eqnarray}
\bigl( 1 \,-\, \frac{v^{2}}{c^{2}} \bigr) x'_{1} &=& L / \gamma \,,
\bigl( 1 \,-\, \frac{v^{2}}{c^{2}} \bigr) x'_{2} &=& 0 \,,
\end{eqnarray}
$$
and since $$1 \,-\, v^{2}/c^{2} \,=\, 1 / \gamma^{2}$$, the boosted trajectories work out to
$$
\begin{eqnarray}
x'_{1} &=& \gamma L \,,
x'_{2} &=& 0 \,.
\end{eqnarray}
$$
As for the distance, $$L' \,=\, x'_{1} \,-\, x'_{2} \,=\, \gamma L$$.
So for this problem, $$L' \,=\, \gamma L$$, like I said right from the beginning.