Misconception about index notation I'm going to give an example in General Relativity but this is a question about index notation and coordinate transformations in general. In "Spacetime and Geometry" by Sean Caroll, there is this definition of the connection
\begin{equation}
\Gamma^{\ \nu'}_{\mu' \ \lambda'}= \Lambda^{\nu'}_{\ \ \ \nu}\  \Lambda^{\mu}_{\ \ \ \mu'}\  \Lambda^{\lambda}_{\ \ \ \lambda'} \ \Gamma^{\ \nu}_{\mu \ \lambda}-\Lambda^{\mu}_{\ \ \ \mu'}\  \Lambda^{\lambda}_{\ \ \ \lambda'}\ \vec{E}_{\lambda}(\Lambda^{\nu'}_{\ \ \ \mu}). \tag{1}
\end{equation}
Where $\Lambda^{\nu'}_{\ \ \ \nu}$ is a change of coordinates from $\{x\}$ to $\{x'\}$. One can multiply everything by many $\Lambda$ matrices and move the last term to the right to obtain 
\begin{equation}
\Gamma^{\ \nu}_{\mu \ \lambda}= \Lambda^{\nu}_{\ \ \ \nu'}\  \Lambda^{\mu'}_{\ \ \ \mu}\  \Lambda^{\lambda'}_{\ \ \ \lambda} \ \Gamma^{\ \nu'}_{\mu' \ \lambda'} + \Lambda^{\nu}_{\ \ \ \nu'}\ \vec{E}_{\lambda}(\Lambda^{\nu'}_{\ \ \ \mu}).\tag{2}
\end{equation}
This equations are clearly different, one gives the Connection written in $\{x\}$ in terms of $\{x'\}$ and the other one does the inverse operation. However, I could have also started by calling one coordinate as the other (naming the primed indeces as unprimed and viceversa). If I would have done this I would have obtain this instead
\begin{equation}
\Gamma^{\ \nu}_{\mu \ \lambda}= \Lambda^{\nu}_{\ \ \ \nu'}\  \Lambda^{\mu'}_{\ \ \ \mu}\  \Lambda^{\lambda'}_{\ \ \ \lambda} \ \Gamma^{\ \nu'}_{\mu' \ \lambda'}-\Lambda^{\mu'}_{\ \ \ \mu}\  \Lambda^{\lambda'}_{\ \ \ \lambda}\ \vec{E}_{\lambda'}(\Lambda^{\nu}_{\ \ \ \mu'}).\tag{3}
\end{equation}
But now the second equation and this equation give different rules for transforming from the primed coordinates to the unprimed ones. Where I am doing something wrong?
 A: *

*OP is considering the transformation formula
$$\frac{\partial y^{\tau}}{\partial x^{\lambda}} \Gamma^{(x)\lambda}_{\mu\nu} ~=~\frac{\partial y^{\rho}}{\partial x^{\mu}}\, \frac{\partial y^{\sigma}}{\partial x^{\nu}}\, \Gamma^{(y)\tau}_{\rho\sigma}+ 
\frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}. \tag{A}
$$
for the Christoffel symbol under general local coordinate transformations $x^{\mu}\to y^{\nu}=y^{\nu}(x)$. OP already knows the Christoffel symbol is not a tensor.

*OP's eqs. (2) and (3) are indeed consistent. One just needs to use the following three facts (transcribed into OP's non-standard notation):
$$ 
 \vec{E}_{\mu}(\Lambda^{\lambda^{\prime}}{}_{\nu})~=~(\mu\leftrightarrow \nu),\qquad 
\vec{E}_{\mu}~=~\Lambda^{\mu^{\prime}}{}_{\mu} \vec{E}_{\mu^{\prime}},\qquad
\vec{E}_{\mu}(\Lambda^{\lambda}{}_{\nu^{\prime}})
~=~-\Lambda^{\lambda}{}_{\lambda^{\prime}} \vec{E}_{\mu}(\Lambda^{\lambda^{\prime}}{}_{\nu}) \Lambda^{\nu}{}_{\nu^{\prime}}. \tag{B}$$
A: In order to not get lost in the debauch of indices, consider the situation of a general $G$-connection, $ \omega_{\mu\ \ b}^{\ a} $. Here $\mu$ is a space-time index, and $a,b$ latin indices are "internal indices".
One may perform a gauge transformation by a $\Lambda^a_{\ b}$ $G$-valued function ($G$ here is a matrix Lie group). The connection transforms under coordinate transformations as a honest covariant vector, but under gauge transformations, it transforms as $$ \omega_{\mu\ \ b}^{'\ a}=(\Lambda^{-1})^a_{\ c}\omega_{\mu\ \ d}^{\ c}\Lambda^d_{\ b}+(\Lambda^{-1})^a_{\ c}\partial_\mu\Lambda^c_{\ b}. $$ To concatenate notation, let us use differential form notation for the space-time index, as in $\omega^a_{\ b}=\omega_{\mu\ \ b}^{\ a}dx^\mu$ and use matrix notation for the internal indices. The transformation law is then $$ \omega'=\Lambda^{-1}\omega\Lambda+\Lambda^{-1}d\Lambda. $$ The "gauge transformation" can be seen as a transformation of "internal" frames. Let us define the initial internal frame as $e$ and the transformed frame as $e'=e\Lambda$. Now, the transformation rule is to be rearranged: $$ \Lambda\omega'\Lambda^{-1}-d\Lambda\Lambda^{-1}=\omega. $$ Now, as you said, we can reverse the "primed" and "unprimed" stuff to get $$ \omega'=\Lambda\omega\Lambda^{-1}-d\Lambda\Lambda^{-1}, $$ and it is right of you to ask the question, which transformation rule is correct.
The resolution is to look at the gauge transformation. Originally it was $$ e'=e\Lambda. $$ If we reverse the role of primed and unprimed objects, we get $$ e=e'\Lambda, $$ which may be rearranged as $$ e\Lambda^{-1}=e'. $$
So you can see that when the role of primed and unprimed objects are reversed, the role of $\Lambda$ and $\Lambda^{-1}$ are also reversed.
Let us check how the "second" transformation rule changes when we change make a change $\Lambda\ ->\ \Lambda^{-1}$:
Obviously $\Lambda\omega\Lambda^{-1}$ goes to $\Lambda^{-1}\omega\Lambda$, so how does the Maurer-Cartan term $-d\Lambda\Lambda^{-1}$ change? It changes to $$ -d(\Lambda^{-1})\Lambda, $$ however because $\Lambda^{-1}\Lambda=1$, we have $$ 0=d1=d(\Lambda^{-1}\Lambda)=d(\Lambda^{-1})\Lambda+\Lambda^{-1}d\Lambda, $$ so we have $$ -d(\Lambda^{-1})\Lambda=\Lambda^{-1}d\Lambda. $$ Therefore, if, during the "primed" $\leftrightarrow$ "unprimed" change, we also incur a $\Lambda\rightarrow\Lambda^{-1}$ change, the "weird" transformation rule $$ \omega'=\Lambda\omega\Lambda^{-1}-d\Lambda\Lambda^{-1} $$ changes to the original transformation rule $$ \omega'=\Lambda^{-1}\omega\Lambda+\Lambda^{-1}d\Lambda. $$
To translate this to GR/Riemannian geometry, we note that for a tangent connection $\Gamma^{\sigma}_{\mu\nu}$, the "internal" and "space-time" indices are the same, and "gauge transformations" are the same as coordinate transformations. Because $$ \partial_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu, $$ we have $$ \Lambda=\frac{\partial x}{\partial x'} $$ with indices suppressed. Of course, it complicates the exact expressions, that gauge transformations cannot be separated from coordinate transformations, so here all three indices transform, but the first lower index transforms as a honest covector.
TL;DR: When you reverse the primes, the coordinates also get reversed, hence the coordinate transformation matrices need to be switched to their inverses. This explains the discrepancy, as your "second" transformation formula is correct, but it looks different because it is expressed with what would be the inverse of the usual transformation matrix.
