Christoffel Symbols Equality Solution? They changed the exercise, so I tried to solve it again:
I have to prove the following:
Let $\Omega \subseteq \mathbb{R}^d$ be open and $g$ a metric field on $\Omega$. For every $\phi \in \mathrm{Diff}(\Omega)$ let $\Xi^i_{jk}[\phi]$ be functions on $\Omega$ that transform in the same way as the Christoffel symbols $\Gamma^i_{jk}[\phi]$
$$
\Xi^i_{jk}[\phi](y) = \frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Xi^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y).
$$
Assume that for each $y_0 \in \Omega$ there is a $\phi_0 \in \mathrm{Diff}(\Omega)$ such that $\Xi^i_{jk}[\phi_0](\phi_0(y_0)) = 0$ and $(\partial_a g_{bc})[\phi_0](\phi_0(y_0))=0$. Show $\Xi^i_{jk}[\phi] =\Gamma^i_{jk}[\phi]$ for all $\phi \in \mathrm{Diff}(\Omega)$.

Here is my try:
We define:
$T^{i}_{\,jk}[\phi](y) := \Xi^i_{\,jk}[\phi](y) - \Gamma^i_{\,jk}[\phi](y)$
Thus we see that this transforms as a Tensor:
$T^{i}_{\,jk}[\phi](y)= \Xi^i_{jk}[\phi](y) - \Gamma^i_{\,jk}[\phi](y) = \bigg(\frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Xi^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y) \bigg)- \bigg(\frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Gamma^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y)\bigg) = \frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y)  (\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y))$
Now we see that for a given pair $y_0$, $\phi_0$:
$\Gamma^i_{\,jk}[\phi_0](\phi_0(y_0))=\frac{1}{2} g^{il}(\partial_k g_{lk} + \partial_j g_{lk} - \partial_l g_{jk})[\phi_0](\phi_0(y_0))= \frac{1}{2} g^{il}\left(\partial_k g_{lk}[\phi_0](\phi_0(y_0)) + \partial_j g_{lk}[\phi_0](\phi_0(y_0)) - \partial_l g_{jk}[\phi_0](\phi_0(y_0))\right)=0$
So for $y_0$ and $\phi_0$ :
$T^{i}_{\,jk}[\phi_0](\phi_0(y_0)) := \Xi^i_{\,jk}[\phi_0](\phi_0(y_0)) - \Gamma^i_{\,jk}[\phi_0](\phi_0(y_0))=0$
Thus
$\frac{\partial x^p}{\partial y^j}(\phi_0(y_0)) \frac{\partial x^q}{\partial y^k}(\phi_0(y_0))  (\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](\phi_0^{-1}(\phi_0(y_0)))\frac{\partial y^i}{\partial x^r}(\phi_0^{-1}(\phi_0(y_0)))= \frac{\partial x^p}{\partial y^j}(\phi_0(y_0))\frac{\partial x^q}{\partial y^k}(\phi_0(y_0))  \left({(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)}\right)\frac{\partial y^i}{\partial x^r}(y_0)\overset{!}{=}0$
Now my BIG question is: can I say that: Because $T^{i}_{\,jk}$ is a Tensor:
$(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)\overset{???}{=}0$
Because IF so then the exercise becomes trivial since:
$T^{i}_{\,jk}[\phi](\phi(y_0))=\frac{\partial x^p}{\partial y^j}(\phi(y_0))\frac{\partial x^q}{\partial y^k}(\phi(y_0))  \overset{\underbrace{=0}}{\left({(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)}\right)}\frac{\partial y^i}{\partial x^r}(y_0)=0$
for all $\phi\in Diff(\Omega)$ and we are done
If not my old proof/idea does not hold anyway and I have got no other clue on how to overcome the problem...
 A: I've finally managed to get a solution which does not imply any assumption which was not given in the statement of task, does it make sense to you guys?
We define:
$T^{i}_{\,jk}[\phi](y) := \Xi^i_{\,jk}[\phi](y) - \Gamma^i_{\,jk}[\phi](y)$
Thus we see that this transforms as a Tensor, hence is a Tensor:
$T^{i}_{\,jk}[\phi](y)= \Xi^i_{jk}[\phi](y) - \Gamma^i_{\,jk}[\phi](y) = \bigg(\frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Xi^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y) \bigg)- \bigg(\frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Gamma^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y)\bigg) = \frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y)  (\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y))$
Now we have that for our given $y_0$ and $\phi_0$:
$\Gamma^i_{\,jk}[\phi_0](\phi_0(y_0))=\frac{1}{2} g^{il}(\partial_k g_{lk} + \partial_j g_{lk} - \partial_l g_{jk})[\phi_0](\phi_0(y_0))= \frac{1}{2} g^{il}\left(\partial_k g_{lk}[\phi_0](\phi_0(y_0)) + \partial_j g_{lk}[\phi_0](\phi_0(y_0)) - \partial_l g_{jk}[\phi_0](\phi_0(y_0))\right)=0$
Now take any $y_0$ and any diffeomorphism $\phi\in \operatorname{Diff}(\Omega)$ :
$T^{i}_{\,jk}[\phi_0](\phi(y_0))=\frac{\partial x^p}{\partial y^j}(\phi(y_0))\frac{\partial x^q}{\partial y^k}(\phi(y_0))  {\left({(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)}\right)}\frac{\partial y^i}{\partial x^r}(y_0)$
By the statement of task we can always find a $\phi_0$ such that:
$T^{i}_{\,jk}[\phi_0](\phi_0(y_0))=0=\frac{\partial x^p}{\partial y^j}(\phi_0(y_0))\frac{\partial x^q}{\partial y^k}(\phi_0(y_0))  {\left({(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)}\right)}\frac{\partial y^i}{\partial x^r}(y_0)$
for each $y_0\in\Omega$.
Since for any diffeomorphism we have that
$\frac{\partial x^p}{\partial y^j}(\phi(y))\frac{\partial x^q}{\partial y^k}(\phi(y))\neq 0$
We see that the transformed part which is invariant under any $\phi\in \operatorname{Diff}(\Omega)$ is the one which maps to zero, we conclude for all of $\Omega$:
$T^i_{jk}[\phi]=\Xi^i_{jk}[\phi]-\Gamma^i_{jk}[\phi]\overset{\forall \phi}{=}0$
$$\Rightarrow \Xi^i_{jk}[\phi]=\Gamma^i_{jk}[\phi]$$
$\blacksquare$
