How to show using Lorentz transformation the $\vec{v}$ of one frame to another in either frame is $\vert\vec{v}\vert$ I am trying to show that in the case of one frame at rest, S, and another frame moving in the +x direction, S', that:


*

*The S frame see's the S' frame moving with $+\vec{v}$

*The S' frame see's the S frame moving with $-\vec{v}$


First, we start in the S frame which see's the S' frame moving in the +x direction. We record two spacetime events of the location of the S' frame's origin and find the velocity:
$$Event 1: (t_1,x_1,0,0) $$
$$Event 2: (t_2,x_2,0,0) $$
$$\frac{x_2 - x_1}{t_2 - t_1}=\frac{\Delta x}{\Delta t}=+\vec{v}$$
Now, for the S' frame, they see the S frame's origin moving away in the -x direction. We can transform the events using time dialation and length contraction arguments:


*

*The length $\Delta x$ reported by the S frame will be longer than the S' frames length (because S' is at rest now), so $\Delta x'=- \frac{\Delta x}{\gamma}$.

*The time interval $\Delta t$ reported by the S frame will be shorter than the S' frames time interval, so $\Delta t'= \Delta t\gamma$.
$$\frac{\Delta x'}{\Delta t'}=\frac{\frac{-\Delta x}{\gamma}}{\Delta t\gamma}  =\frac{-v}{\gamma^2}$$
Which is obviously wrong, but I am really stuck on how to set up this problem/system.
 A: To measure length contraction you need to look at two parallel world-lines; you are looking at one individual world-line and therefore you've not been clear about what you're studying.
The coordinate transformation that you want is (with $w=ct$ and $\beta=v/c$ to show a certain symmetry of the equations)$$\begin{align}
w'&=\gamma~(w - \beta~x),\\
x'&=\gamma~(x - \beta~w),\\
y'&=y,\\
z'&=z.\end{align}$$
Ignoring the $y$ and $z$ directions as trivial, length contraction comes when we consider the world-lines $(w, x) = (s, 0)$ for all $s$ and $(w, x)=(s, L)$ for all $s$. These both transform to:$$w'=\gamma~(s - \beta~L),\\
x'=\gamma~(L-\beta~s).$$Substituting for $s=w'/\gamma + \beta~L$ we can find the equation $$x'=\gamma~(L-\beta~(w'/\gamma+\beta~L) = -\beta~w' + \gamma~L~(1 - \beta^2).$$
Note that $\gamma~(1-\beta^2) = \gamma / \gamma^2 = 1/\gamma,$ giving our familiar length contraction factor; note also that this line which was stationary is now moving backwards with speed $-\beta~c=-v$.
This effect that when you accelerate, clocks in front of you tick faster the further they are in front of you, and clocks behind you tick slower the further they are behind you, is called the "relativity of simultaneity." It is a very important effect: in fact I have argued before that it is the most important effect, in that for small $\beta\ll 1$ it is the only effect which survives: and we can reconstruct the general coordinate transformation above from this $\gamma\approx 1$ form. 
A: You speak of relativistic reciprocity, i.e. the assertion that the inverse Lorentz transformation is found by making the substitution $v\mapsto-v$, as discussed further in my answer here. 
Although intuitively reasonable and deceptively simple, it is not trivial and it must either be taken as a postulate itself or it can be derived from some axiomatic beginnings for special relativity. 
The essence of the matter is an assumption of isotropy of space: that if we change the direction of a boost and re-align our co-ordinate system to the boost, the Lorentz transformation cannot change. That is, there is no special or preferred direction in space and all directions "look the same". Step 4 in my detailed explanation below is the crucial and most relevant-to-the-problem at hand way that the spatial isotropy assumption enters the  discussion.
But the number of assumptions you need to make to reach the point you are trying to prove is remarkably many. You can refer to the paper linked in my other answer, or my own take on the matter is sketched in the details below.

Detailed Necessary Axioms and Derivation of Lorentz Transformation, with Reciprocity


*

*The four axioms of 


*

*(1) Galileo's principle (that the Lorentz transformation depends only on relative velocity), together with assumptions of 

*(2) spacetime homogeneity;

*(3) spacetime flatness (or of a spacetime manifold and that we are dealing with local transformations so that we can at least have local flatness) and 

*(4) continuity of the Lorentz transformation (in its dependence on spacetime co-ordinates) shows that the Lorentz transformation acts linearly on spacetime co-ordinates, i.e. can be described by a matrix;


*A fifth axiom that the Lorentz transformation in a given direction maps the real line of velocities continuously into a group of matrices shows that the transformation must be of the form: $$X\mapsto \exp(\eta(v)\,K)\,X\tag{1}$$ where the rapidity $\eta(v)$ is a continuous function of the velocity $v$, $K$ is a constant $4\times 4$ matrix and $X$ the $1\times 4$ column vector of spacetime co-ordinates. I show this in greater detail in my answer here.

*Now we think about isotropy of space. If space is isotropic, we are free to choose the $x$ axis to be along the direction of the boost and we cannot lose generalness. A boost in any other direction is simply found by rotation of co-ordinates from the form we calculate in the $x$ direction. Moreover, an arbitrary rotation of co-ordinates about the $x$ axis cannot change the form of our boost either. Thus, from (1): $$R_x(\theta)\exp(\eta\,K)\, R_x(\theta)^{-1} = \exp(\eta\,K)\tag{2} $$ where $R_x(\theta)$ is the rotation through angle $\theta$ about $x$. If our co-ordinate order is $(t,\,x,y,\,z)$ then (2) can be shown to reduce the matrix $K$ to the form: $$K = \left(\begin{array}{cc}k_{11}&k_{12}&0&0\\k_{21}&k_{22}&0&0\\0&0&k_{33}&-k_{43}\\0&0&k_{43}&k_{33}\end{array}\right)\tag{3}$$

*A second, a crucial to your question, isotropy of space consideration is what happens when we reverse the co-ordinate axis in the direction of the boost. If we reverse one of the other co-ordinate axes, so that the spatial co-ordinates stay right handed, then an isotropic space means that the transformation cannot change. Our co-ordinate transformation matrix is then, say, $\mathrm{diag}(1,\,-1,\,-1,\,1)$, which is a $180^\circ$ rotation about the $z$ axis (the same result follows from a $180^\circ$ rotation about any axis normal to the boost direction). So we have the condition: $$\eta \,\mathrm{diag}(1,\,-1,\,-1,\,1)\,K\,\mathrm{diag}(1,\,-1,\,-1,\,1) = \eta^\prime\, K\tag{3}$$ That is, the boost is in the same direction, i.e. represents "the same kind of motion", thus is still of the form (1) with a possibly different rapidity parameter $\eta^\prime$. This leads to the equation: $$\left(\begin{array}{cc}(1+u)\,k_{11}&-(1-u)\,k_{12}&0&0\\-(1-u)\,k_{21}&(1+u)\,k_{22}&0&0\\0&0&(1+u)\,k_{33}&(1-u)\,k_{43}\\0&0&-(1-u)\,k_{43}&(1+u)\,k_{33}\end{array}\right)=0\tag{4}$$ where $u=\eta^\prime/\eta$. There are now two possibilities, $u=\pm1$, for a nontrivial (i.e. $K_{jk}=0;\;\forall j,\,k$) Lorentz transformation. The first is $\eta=\eta^\prime$. There's nothing in the postulates so far that forbids this, but it would be a weird universe wherein such a law held. The Lorentz transformation would be diagonal, and boosts would not be relative motion at all; indeed, instead of relative motion, we'd have time dilation together with shrinking and swelling of length scales. This is what Sean Carroll calls the "Alice in Wonderland" universe; where beings can't really move, but can shrink and stretch at will, like Alice drinking her buttered toast "Drink Me" and nibbling the edges of her mushroom. The second possibility is $u=-1$. This is essentially what you want to prove, i.e. that the same motion of the same rapidity but in the opposite direction corresponds to $\eta\mapsto-\eta$ and to the inverse Lorentz transformation. So we're essentially done, but if we want to see the same result cast in terms of velocities, we focus on the $(x,\,t)$ co-ordinates, for which the transformation is: $$\begin{array}{lcl}\left(\begin{array}{c}t\\x\end{array}\right)&\mapsto&\exp\left(\eta\left(\begin{array}{cc}0&k_{12}\\k_{21}&0\end{array}\right)\right)\,\left(\begin{array}{c}t\\x\end{array}\right)\\\\&=&\left(\begin{array}{cc}\cosh(\sqrt{k_{12}\,k_{21}}\,\eta)&\sqrt{\frac{k_{12}}{k_{21}}}\sinh(\sqrt{k_{12}\,k_{21}}\,\eta)\\\sqrt{\frac{k_{21}}{k_{12}}}\sinh(\sqrt{k_{12}\,k_{21}}\,\eta)&\cosh(\sqrt{k_{12}\,k_{21}}\,\eta)\end{array}\right)\,\left(\begin{array}{c}t\\x\end{array}\right)\end{array}\tag{5}$$ which we can now "calibrate" to work out the relationship between velocity $v$ and rapidity $\eta(v)$ by finding the relationship that sets $x=0$ in the transformed co-ordinates: $$v =  \sqrt{\frac{k_{21}}{k_{12}}}\tanh(\sqrt{k_{12}\,k_{21}}\,\eta)\tag{6}$$ which is an odd function in $v$, showing that the inverse Lorentz transformation is found by making the substitution $v\mapsto-v$, which is what you set out to prove. 


Note that this is nontrivial and nonobvious, particularly because there is the Alice in Wonderland universe that is compatible with all our assumptions including spatial isotropy, so we need, further to our other axioms, an explicit ruling out of this possibility as an axiom / assumption. The Alice in Wonderland universe has the same (rather than inverse) transformation wrought by direction reversal.

Afterword: Tidying Up - Causality and the Last Steps to the Usual Lorentz Transformation
To deal with the $k_{43}$ in the above, i.e. to prove $k_{43}=0$, i.e. the curious twist about the $x$ axis that is allowed by our assumptions so far, we need to make further assumptions; either to explicitly postulate $k_{43}=0$ on the experimental grounds that this twist is not observed experimentally or, alternatively, an assumption that the reversal of the $t$ axis corresponds to the inverse Lorentz transformation will also prove the same thing. Lastly, one has to investigate the sign of $k_{12}\,k_{21}$; if negative, the Lorentz transformation is a rotation, which is hard to reconcile with causality. So we must have that $k_{12}\,k_{21}>0$, which, on a suitable change of units, leads to the familiar Lorentz boost in (5)
