Are constant connection coefficients uniquely determined by the (1,3) curvature coefficients? Suppose that on a certain coordinate system the coefficients $\Gamma^i_{jk}$, $i,j,k=1,\cdots, n$, of a linear connection are constant. We do not require compatibility with a metric, however I am interested in the torsionless totally symmetric case $\Gamma^i_{jk}=\Gamma^k_{ij}=\Gamma^i_{kj}$, so that $\Gamma$ has  $\begin{pmatrix}  n+2\\ 3 \end{pmatrix}=\frac{1}{6}(n+2)(n+1) n$ components (the succession for $n\ge 2$ is 4, 10, 20, 35 $\cdots$) while the curvature (I use Einstein's sum convention)
$$ 
R^i_{jkl}= \Gamma^i_{m k} \Gamma^m_{jl}-\Gamma^i_{ml} \Gamma^m_{jk}. 
$$
has  $\frac{1}{12} n^2 (n^2-1)$ components (the succession is 1, 6, 20, 50 $\cdots$). Thus for $n\ge 4$ there are as many curvature constraints as connection components, and actually more constraints than components for $n>4$. Can there be for $n\ge 4$ two sets of connection coefficients $\{\Gamma^i_{jk}\}$, $\{\tilde \Gamma^i_{jk}\}$, saved for the obvious possibility $\tilde\Gamma=-\Gamma$, with the same curvature coefficients as given in the above displayed equation? What if they are traceless? Clearly, this is a quadratic tensor equation, I mentioned the curvature tensor just to motivate the interest in this type of equations. Do you have any suggestions on how to solve this type of overdetermined problems?
I should probably add my attempted strategy. 
I had in mind a possible solution for each $n\ge 4$. Essentially we have here a (homogeneous) quadratic map $f\colon \mathbb{R}^p \to \mathbb{R}^q$, where $p$ is the number of connection coefficients and $q$ the number of curvature coefficients. For $n\ge 4$, $q\ge p$. Each component $f^i$ is quadratic. For quadratic map the following mean value equation holds $f(y)-f(x)=Jf(\frac{x+y}{2}) (y-x)$ thus if $\textrm{Rank} Jf(v) =p$ for $v\ne 0$ then $x$ and $y$ cannot have the same image unless $y=-x$. Thus to confirm injectivity we have only to check $\textrm{Ker}\, Jf(v) =\emptyset$ for $v\ne 0$ (which for $n=4$ means to check $\det Jf\ne 0$). This can be done for any given $n$, the only problem is that it is very time consuming and does not prove injectivity for every $n\ge 4$. For instance $Jf$ for $n=4$ is a $20 \times 20$ matrix (I did not try to calculate it). 
 A: It's not hard to see that you don't always have uniqueness up to sign when $n\ge 4$.  Just consider the case when $\Gamma^i_{jk}$ is zero except when $i=j=k$.  In this case, one has $R^i_{jkl}=0$ for all $i$, $j$, $k$, and $l$, independent of the values of $\Gamma^i_{ii}$, so, in dimension $n$, this gives an $n$-parameter family of solutions to the equations $R^i_{jkl}=0$. (This is not the general solution, though, even for $R=0$. Using the invariance of the equations under $\mathrm{O}(n)$, one sees that the space of solutions to $R=0$ in dimension $n$ is a variety of dimension at least $\tfrac12n(n{+}1)$.)
However, the case $R^i_{jkl}=0$ is very far from the 'generic' case.
You can see that this is not an isolated problem with $R=0$ by considering the following example in dimension~$4$.  Let $b_1$, $b_2$, and $b_3$ be constants satisfying ${b_1}^2-{b_2}^2-{b_3}^2=0$ and set
$$
\begin{aligned}
C &= \left( b_{3}{+}b_{1} \right)  \left( x_{1}{+}x_{2}{+}x_{3}{+}x_{4}\right)^{3}\\
&\qquad+\ 6\left( b_{{2}}{-}b_{{3}}{-}b_{{1}}\right) 
 \left(  x_{1}{+}x_{2}{+}x_{3}{+}x_{4} \right)\left(
 {x_{1}}^{2}{+}{x_{2}}^{2}{+}{x_{3}}^{2}{+}{x_{4}}^{2}\right)\\
&\qquad\qquad +\ 8\,b_{{3}} 
\left( {x_{1}}^{3}+{x_{2}}^{3}+{x_{3}}^{3}+{x_{4}}^{3} \right).
\end{aligned} 
$$
Now, define 
$$
\Gamma^i_{jk} = \frac1{24}\frac{\partial^3C}{\partial x^i\partial x^j\partial x^k}.
$$
Then one easily computes that $R^i_{jkl} = 2b_2(b_1-b_2)\left(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}\right)$, i.e., the curvature tensor $R$ has constant sectional curvature $2b_2(b_1-b_2)$.  Note, by the way, that this connection will be traceless if $b_1 = \tfrac32 b_2$, in which case the sectional curvature will be ${b_2}^2$. 
This gives a couple of $1$-parameter families of connections with the same constant sectional curvature $s$ for each $s$ in $\mathbb{R}$, and a traceless example for each $s\ge0$.  Now, acting by the orthogonal group $\mathrm{O}(4)$, you see that each of these curvatures (which are invariant under $\mathrm{O}(4)$) comes from a space of connections (respectively, traceless connections) that has dimension at least $7$ (respectively, $6$), since the $O(4)$ symmetry group of $C$ is discrete for most choices of the $b_i$.  Thus, you cannot hope to have uniqueness, even though it's still possible that there is some open set of curvature tensors in dimension $4$ that come from only a finite number of constant coefficient connections.  
This nonunqiueness certainly persists in dimensions above $4$.  In fact, the analogous $3$-parameter family of $C$s invariant under the symmetric group $S_n$ that you can write down in any dimension $n\ge4$ furnish counterexamples (many traceless) to the desired quadratic uniqueness when the target $R$ has constant sectional curvature, i.e., is of the form $R = s(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})$.
Probably, your best bet, in the generic case, is to use the Bianchi identities to reduce the problem:  If $\omega = (\omega^i_j)$ where $\omega^i_j = \Gamma^i_{jk}\,\mathrm{d}x^k$, then $\omega$ is a constant coefficient $1$-form with values in the symmetric $n$-by-$n$ matrices.  Your curvature equation is
$$
\omega\wedge\omega = \Omega\tag1
$$
where $\Omega = (\Omega^i_j)$ satisfies $\Omega^i_j = {\tfrac12} R^i_{jkl}\,\mathrm{d}x^k\wedge\mathrm{d}x^l$ and $\Omega$ is, of course, skewsymmetric and closed.  
Now, regard $\Omega$ as known and solve for $\omega$. To simplify this one can use the 'Bianchi relation'
$$
\Omega\wedge\omega - \omega\wedge\Omega = \mathrm{d}\Omega = 0,\tag2
$$
a triviality in this case.  The system $(2)$ is a linear set of equations for $\omega$ with coefficients that are linear in the $R^i_{jkl}$, and restricting to the solutions of this system will usually simplify the search for the solutions to the original quadratic equations $(1)$.  
For the generic choice of $R^i_{jkl}$ satisfying the curvature identities, when $n$ is sufficiently large, $(2)$ is an overdetermined linear system of equations for the coefficients of the symmetric matrix $\omega$, and, for generic curvature tensor~$R$, it will have $\omega=0$ as the only solution.  In such case, there will be no solution to $(1)$.  In general, one can use $(2)$ to stratify $R$-space according to the rank of the equations $(2)$, and then one can work on the strata of constant rank.
Added remark in dimension $4$:  It is not hard to write down an explicit $R$ in dimension $4$ such that the only solution to the linear system $(2)$ is $\omega=0$.  Such  curvatures $R$ form a nonempty Zariski-open set $U\subset\mathbb{R}^{20}$, and, obviously, no such $R$ is the curvature of a symmetric constant connection in the OP's sense.  In particular, it follows that the image of the OP's map $f:\mathbb{R}^{20}\to\mathbb{R}^{20}$ must lie in the proper Zariski-closed set that is the complement of $U$ and hence must have strictly smaller dimension than $20$.  It follows that the Jacobian of $f$ at every point of its domain must have rank strictly smaller than $20$, and hence that the generic nonempty fiber of $f$ must have dimension strictly greater than $0$.  Thus, the OP's desired 'uniqueness up to sign' fails completely when $n=4$!
In the next lower 'stratum' in $R$-space, the space of solutions to $(2)$ will probably have dimension $1$. Suppose, it is generated by some $\omega_0$.  Then you have to look at the equations $(t\omega_0)\wedge(t\omega_0)-\Omega = 0$ and hope that this quadratic equation for $t$ has a solution.  Generically, in this stratum, it won't, when $n$ is sufficiently large, but, when it does, this will provide the only solution.
Generally, what you should do is look at the vector space $S(\Omega)$ of solutions $\omega$ to $(2)$ and then regard $(1)$ as a system of quadratic equations for elements of $S(\Omega)$.  For most $\Omega$, when $n\ge 4$, this reduces it to a manageable problem, though, most of the time, there are no solutions.
Further remarks in dimension $4$:  Calculation shows that, when $n=4$ the maximum rank of the Jacobian of the map $f:\mathbb{R}^{20}\to\mathbb{R}^{20}$ described by the OP above is $18$.  This implies both that the closure of the image of $f$ is an algebraic variety of dimension $18$ in the range $\mathbb{R}^{20}$ and that the generic element $v$ in the domain of $f$ lies in an $f$-fiber of dimension $2$, so nonuniqueness fails generically in dimension $4$, in spite of a reasonable expectation of at least finiteness based on a simple dimension count.  Moreover, for generic $v$ in the domain of $f$, it turns out that the corresponding linear system $(2)$ has a solution space $S(\Omega)$ of dimension $4$.  This means that, in order to find all of the solutions to $\omega\wedge\omega = \omega_0\wedge\omega_0$ when $\omega_0$ is generic, one only has to solve the quadratic equations $\omega\wedge\omega = \omega_0\wedge\omega_0$ for $\omega$ lying in a vector space of dimension $4$ (not $20$), and, moreover one knows that the space of solutions will be a surface in that $4$-dimensional vector space.  Thus, this drastically simplifies the problem of computing the fibers of $f$ for a generic curvature that is known to be in the image of $f$.  (I admit that I am surprised that the dimension of $S(\omega_0\wedge\omega_0)$ is generically $4$ rather than $2$.  This is another mysterious feature of this algebra problem that I find intriguing.)
This problem is also interesting for $n=3$, but, in that case, one can use the representation theory of $\mathrm{O}(3)$ to reduce the problem to a manageable algebra problem.  In fact, one finds that, when $R$ is a curvature tensor of nonconstant sectional curvature, then there is a $4$-parameter family of solutions $\Gamma$, the expected dimension.  However, when $R$ falls on the line of curvature tensors that have constant sectional curvature, it turns out that the space of solutions $\Gamma$ has dimension $6$.  (Interestingly enough, it is a (singular) rational variety in all cases!)
