Four-vector and Notation significance As the title suggest, this has to do, on the most part, with four vector notation. I have a series of questions, the majority, related to this topic:
1- If we assume a lorentz boost in the x direction then, is my vector/matrix representation for the contravariant and covariant four vectors correct?:
$x^{ ' \mu}=\sum_v \Lambda_\nu^{\;\mu}x^\nu=(x_0',x_1',x_2',x_3')\Lambda_\nu^{\;\mu}(x_0,x_1,x_2,x_3)$
$x_\mu^{'}=\sum_v \Lambda_\mu^{\;\nu}x_\nu=(x_0',-x_1',-x_2',-x_3')\Lambda_\nu^{\;\mu}(x_0,-x_1,-x_2,-x_3)$
The LT in the 2nd equation is the inverse. But I don't know how the indices between two matrices, where one is the inverse of the other, stand or what they look like. This will be one of the below asked questions.
2 - Is there a difference between: $\Lambda_\mu^{\;\nu}$ and $\Lambda_{\; \mu}^{\nu}$?
3 - Should contracted indices always stand one after the other: $x^{' \mu}= \Lambda_\nu^{\;\mu}x^\nu$ or $x^{' \mu}= \Lambda_{\; \nu}^{\mu} x^{\nu}$ ?
4- From the inner product between two four vectors the following equation can be reached:
$g_{\mu \nu}\Lambda_{\alpha}^{\; \mu}\Lambda_{\beta}^{\; \nu}=g_{\alpha \beta}$
$\Lambda_{\alpha}^{\; \mu}g_{\mu \nu} \Lambda_{\beta}^{\; \nu}=g_{\alpha \beta}$
Can one, swap components/elements like above, without no problem at all?Or is it valid only for the metric tensor?
5 - Is it correct to assume: relativistic invariant = Lorentz invariant ?
6 - If $\Lambda_{\rho}^{\; \mu}$ is a Lorentz Transformation, then the inverse of it, how is it represented? Like this: $\Lambda_{\; \rho}^{\mu}$. This means:
$$(\Lambda^{-1})_{\rho}^{\; \mu}= \Lambda_{\; \rho}^{\mu}$$ ?
 A: 
1- If we assume a lorentz boost in the x direction then, is my vector/matrix representation for the contravariant and covariant four vectors correct? $$ x^{ ' \mu}=\sum_v \Lambda_\nu^{\;\mu}x^\nu=(x_0',x_1',x_2',x_3')\Lambda_\nu^{\;\mu}(x_0,x_1,x_2,x_3) \\ x_\mu^{'}=\sum_v \Lambda_\mu^{\;\nu}x_\nu=(x_0',-x_1',-x_2',-x_3')\Lambda_\nu^{\;\mu}(x_0,-x_1,-x_2,-x_3)$$

It is not entirely clear what you've done here. In fact, your expression on the RHS makes no sense, immediately from the summation. What we have is essentially,
$$ x'^{\mu}=\sum_\nu \Lambda_\nu^{\;\mu}x^\nu = \Lambda_0^{\;\mu}x^0 + \Lambda_1^{\;\mu}x^1 + \Lambda_2^{\;\mu}x^2 + \Lambda_3^{\;\mu}x^3, \\ x'_\mu =\sum_\nu \Lambda_\mu^{\;\nu}x_\nu = \Lambda_\mu^{\;0}x_0 + \Lambda_\mu^{\;1}x_1 + \Lambda_\mu^{\;2}x_2 + \Lambda_\mu^{\;3}x_3. $$
Note: $x^\mu$ (and equivalently $x_\mu$) are the $\mu^{\textrm{th}}$ components of the four-vector and not the vector itself.
As you can see, this is a regular summation over the indices. There are no negatives popping out of anywhere. However, they can pop-up if you use the expression, $x_{\mu} = \eta_{\mu \nu}x^\nu$, where, $\eta = \mathrm{diag}(+1,-1,-1,-1)$ is the Minkowski metric in Cartesian co-ordinates. Using the summation convention, we have,
$$ x'_\mu = \Lambda_\mu^{\;\nu} x_\nu = \Lambda_\mu^{\;\nu} \eta_{\nu \sigma} x^\sigma = \Lambda_\mu^{\;0}x^0 - \Lambda_\mu^{\;1}x^1 - \Lambda_\mu^{\;2}x^2 - \Lambda_\mu^{\;3}x^3. $$
For a more general metric $g$ however, one requires more than a simple change in sign, so this does not hold. The Minkowski metric is the only one you need to care about in Special Relativity, but more general metrics are required while dealing with General Relativity, where, $x_{\mu} = g_{\mu \nu}x^\nu$ is the expression for lowering indices.

2 - Is there a difference between: $\Lambda_\mu^{\;\nu}$ and $\Lambda^\nu_{\;\mu}$?

Indeed, there is. In the first expression, the component $\Lambda_\mu^{\;\nu}$ coincides with the $\mu^{\textrm{th}}$ row and $\nu^{\textrm{th}}$ column of the transformation matrix $\Lambda$. On the other hand, $\Lambda^\nu_{\;\mu}$ coincides with the $\nu^{\textrm{th}}$ row and $\mu^{\textrm{th}}$ column. But the component $\Lambda_\mu^{\;\nu}$ corresponds to the $\nu^{\textrm{th}}$ row and $\mu^{\textrm{th}}$ column of the transpose $\Lambda^T$, i.e, $\Lambda_\mu^{\;\nu} = (\Lambda^T)^\nu_{\;\mu}$.

3 - Should contracted indices always stand one after the other: $x'^{\mu} =  \Lambda_\nu^{\;\mu}x^\nu$ or $x'^{\mu} = \Lambda_{\;\nu}^{\mu}x^\nu$?

In simple matrix terminology, one just has to remember that row indices contract column indices. As we saw earlier, the first index in a matrix corresponds to a row index and we know that the indices of a contravariant vector are column indices. And since we now know how transposes work, both expressions are valid, but represent different things. In matrix notation, we have,
$$ x'^{\mu} = \Lambda_\nu^{\;\mu}x^\nu \implies x' = \Lambda x, \\ x'^{\mu} =  \Lambda_{\;\nu}^{\mu}x^\nu \implies x' = \Lambda^T x.$$

4- $\dots$ Can one, swap components/elements like above, without no problem at all?Or is it valid only for the metric tensor?

The swapping of components around is generally possible in index notation as each component is a scalar, so the multiplication is commutative. However, one must keep in mind that there is an implied summation. It can easily be checked that the final expressions remain the same upon expanding the summation over indices. Note however that the property mentioned in your example is specific only to the metric, while the shuffling around of components is a general property.

5- Is it correct to assume: relativistic invariant = Lorentz invariant ?

In most classical scenarios this appears to be true, but the actual relativistic invariance is equivalent to Poincaré invariance, where the relevant Poincaré transformations are essentially a combination of Lorentz transformations and space-time translations.

6- $\dots$

As for your last query, I leave you with the expression, $\Lambda^T \eta \Lambda = \eta$, where $\eta$ is the Minkowski metric. Using this one can generate an expression for $\Lambda^{-1}$ and find its relation to $\Lambda$ in index notation. I leave this to you as an exercise to better your understanding.
Hope this helps :)
