This is something I have been thinking about recently, allow me to complete Mariano Suárez-Alvarez' answer.
First just an observation: "pick any Riemannian manifold with trivial holonomy at each point: for example, a space form of curvature zero": actually you have no choice, a connection with trivial holonomy has vanishing curvature, so the only candidate metrics are Euclidean. In the language of geometric structures, a flat torsion-free connection is equivalent to an affine structure.
Now to the point. As we are going to see, the answer to your question is "generically, yes". First note that given a connection $\nabla$ (let's permanently assume that $\nabla$ is torsion-free otherwise there's no chance of it being a Riemanniann connection), by definition a metric $g$ has Levi-Civita connection $\nabla$ if and only if $g$ is $\nabla$-parallel: $\nabla g = 0$.
Let us make a general observation about parallel tensor fields with respect to a given connection. If $F$ is a parallel tensor field, then $F$ is preserved by parallel transport. In particular:
- $F$ is completely determined by what it is at some point $x_0 \in M$ (to find $F_x$, parallel transport $F_{x_0}$ along some path from $x_0$ to $x$).
- $F_{x_0}$ must be invariant under the holonomy group $\operatorname{Hol}(\nabla, x_0)$.
Conversely, given a tensor $F_{x_0}$ in some tangent space $T_{x_0} M$ such that $F_{x_0}$ is invariant under $\operatorname{Hol}(\nabla, x_0)$, there is a unique parallel tensor field $F$ on $M$ extending $F_{x_0}$ (obtained by parallel-transporting $F_{x_0}$).
So you have the answer to your question in the following form:
There are as many Riemannian metrics having Levi-Civita connection
$\nabla$ as there are inner products $g$ in $T_{x_0} M$ preserved by
$\operatorname{Hol}(\nabla, x_0)$.
Now you may want to push the analysis further: how many is that? The answer is provided by analysing the action of the restricted holonomy group $\operatorname{Hol}_0(\nabla, x_0)$ on $T_{x_0}M$. Now this is just linear algebra: let's just call $G = \operatorname{Hol}_0(\nabla, x_0)$ and $V = T_{x_0} M$. Let $g$ and $h$ be two inner products in $V$ that are preserved by $G$, in other words $G \subset O(g)$ and $G \subset O(h)$. If $G$ acts irreducibly on $V$, i.e. there are no $G$-stable subspaces $\{0\} \subsetneq W \subsetneq V$, then a little exercise that I am leaving to you shows that $g$ and $h$ must be proportional. So in the generic case where $\nabla$ is irreducible, the answer to your question is yes:
If $\nabla$ is irreducible, all Riemannian metrics with connection $\nabla$ must be equal up to
positive scalars.
NB: note that there might not be any such metrics if $G$ does not preserve any inner product on $V$, in other words $G$ must be conjugated to a subgroup of $O(n)$.
On the opposite side of the spectrum, if $G$ is trivial, i.e. $\nabla$ is flat, then $g$ and $h$ can be anything, there are no restrictions:
If $\nabla$ is flat, there are as many Riemannian metrics with
connection $\nabla$ as there are inner products in a $\dim M$-dimensional vector space, they are the Euclidean metrics on $M$.
In the "general" case where $\nabla$ is reducible, I hope I am not mistaken (I won't write the details) in saying that you can derive from the de Rham decomposition theorem that the situation is a mix of the two previous "extreme" cases:
If $\nabla$ is reducible, locally one can write $M = M_0 \times N$,
such that both $g$ and $h$ split as products. The components of $g$ and $h$ on $M_0$ are both Euclidean and their components on $N$ are equal up to a scalar.
If this is correct, I believe your question is answered completely.
NB: In this paper (see also this one), Richard Atkins addresses this question. I haven't really looked but since it seems to me that there is not much more to say than what I've written, I have no idea what he's really doing in there.
If $f:(S,g)\to (S',g')$ is an isometry, then define
$\nabla_{X'}Y':=df\ \nabla_XY$
Show that this is LC-connection :
(1) Compatibility condition : First show that $$ X'(Y',Z')=X(Y,Z)$$
Proof : If $\frac{d}{dt}p(t)=X,\ p(0)=p$ then
$$ df_p X(df_p Y, df_p Z) =\frac{d}{dt} (df Y, df Z)_{f(p(t))} =
\frac{d}{dt} (Y,Z)_{p(t)}
$$ since $f$ is an isometry And $\frac{d}{dt} (Y,Z)_{p(t)}= X(Y,Z)$
So $$ (\nabla_{X'}Y',Z')+(Y',\nabla_{X'}Z')=f^\ast g'( \nabla_XY,Z)
+ f^\ast g' (Y,\nabla_XZ) = X(Y,Z) =X'(Y',Z') $$
(2) Symmetry condition : $$ \nabla_{X'}Y' -\nabla_{Y'}X'=df
(\nabla_XY-\nabla_YX)=df[X,Y]=[X',Y']$$
Best Answer
Let $\nabla$ and $\nabla'$ be two affine connections on the same manifold. As you point out, there is a $(1,2)$ tensor $T$ such that $\nabla'_XY=\nabla_XY+T(X,Y)$. Direct computation shows that their respective torsion tensors $\tau,\tau'$ are related by $$ \tau'(X,Y)=\tau(X,Y)+T(X,Y)-T(Y,X) $$ Thus, if $\nabla$ is torsion free, $\nabla'$ is also torsion free iff $T$ is symmetric (in its covariant components).