Can an arbitrary curve locally be made into a geodesic through a conformal change of the metric? 
Given a nowhere null curve $\gamma: I \rightarrow M$ in a pseudo-Riemannian manifold $(M,g)$. (I is some open interval)
  Let $t_0 \in I$ arbitrary.
  Does a conformally equivalent metric $\hat{g}=e^{2\sigma} g$ and some $\epsilon >0$ exist, such that $\gamma \mid _{(t_0-\epsilon, t_0 + \epsilon)}$ is a geodesic in $(M,\hat{g})$?

$\gamma$ being a geodesic with respect to $(M, \hat{g})$ means
$$0=\hat{\nabla}_{\gamma'}\gamma'=\nabla_{\gamma'}\gamma'+ 2 \gamma' \cdot \gamma'(\sigma) - g(\gamma', \gamma') \cdot \operatorname{grad} \sigma,$$
where we used the transformation law for the Levi Civita connection under a conformal change of metrics.
Now it feels there should be some $\sigma$ satisfying this equation locally, because we have very big freedom.
However I am having difficulties carrying out the construction of $\sigma$ and I don't know of general fact that would give existence of a solution.
 A: Interesting question. The answer seems to be yes. (The problem with @JohnHughes's counterexample is that his curve $\gamma$ does not satisfy $\nabla_{\gamma'}\gamma'=0$ with respect to his original metric $g$. If my back-of-the envelope computation is correct, it should be $\nabla_{\gamma'}\gamma'=\frac12 \partial/\partial y$.)
Here's a sketch of a proof that it's always possible to conformally change the metric in a neighborhood of a point to make $\gamma$ a geodesic there. The proof comes in several steps. (Note that all of these constructions are merely local.)
Step 1. Choose local coordinates $(x^1,\dots,x^n)$ in which $\gamma(t) = (t,0,\dots,0)$.  Because we're assuming $\gamma'(t)$ is never zero, this is possible by the rank theorem. 
Step 2. Make a preliminary conformal change to $g$ to make $\gamma$ a unit-speed curve. In the coordinates above, $|\gamma'(t)|^2_g = g_{11}(t,0,\dots,0)$, so this can be accomplished by defining $\overline  g = f g$, where $f(x^1,\dots,x^n) = 1/ g_{11}(x^1,0,\dots,0)$. Let's replace the original metric by $\overline {g}$. 
Step 3. Now change $g$ again to make $\hat \nabla_{\gamma'}\gamma'=0$. Write the coordinate representation of $\nabla_{\gamma'(t)}\gamma'(t)$ as $\sum_j a^j(t)\partial_j|_{\gamma(t)}$. Differentiating the equation $|\gamma'(t)|^2 \equiv 1$ shows that $\nabla_{\gamma'}\gamma'$ is orthogonal to $\gamma'$ along $\gamma$, which means that 
$$
\sum_j g_{1j}(t,0,\dots,0) a^j(t) \equiv 0.\tag{1}
$$
Define a function $\sigma$ by
$$
\sigma(x^1,\dots,x^n) = 1 + \sum_{i,j=1}^n g_{ij}(x^1,0,\dots,0) x^i a^j(x^1) 
 = 1 + \sum_{i=2}^n\sum_{j=1}^n g_{ij}(x^1,0,\dots,0)x^i a^j(x^1) ,
$$
where the second equality follows from $(1)$, and set $\hat g = e^{2\sigma}g$.
Note that along the image of $\gamma$, where $x^2=\dots =x^n=0$, we have $\sigma\equiv 1$, so $\gamma$ is still unit-speed with respect to $\hat\gamma$. Also along the image of $\gamma$, we have $\partial_i \sigma = \sum_j g_{ij}a^j$: for $i=1$, this follows by noting that $\partial_1\sigma=0$ since $\sigma=1$ there, and applying $(1)$; while for $i>1$, it follows by differentiating the formula for $\sigma$ and setting $x^2=\dots= x^n=0$. Therefore,
$$
\operatorname{grad} \sigma = \sum_{i,k} g^{ik} \partial_i\sigma \partial_k = \sum_{i,j,k} g^{ik} g_{ij} a^j \partial_k =\sum_j a^j \partial_j =  \nabla_{\gamma'}{\gamma'}.
$$
Moreover, we also have $\gamma'(\sigma) = \partial\sigma/\partial x^1 = 0$ and $g(\gamma',\gamma') = 1$ along $\gamma$, so it follows from the OP's formula that $\hat  \nabla_{\gamma'}\gamma' \equiv 0$.  $\square$
There might be a more efficient way to do this, but this was the best I could come up with on the spur of the moment.
A: Following Jack Lee's interpretation, I think that the answer is "no". 
Let's look at the case where $M$ is the upper half-plane, and the metric at location $(x, y)$ is given, in the standard basis, by 
\begin{align}
g(x, y) = \begin{bmatrix} 1 & 0 \\ 0 & y \end{bmatrix}
\end{align}
And let's also look at the path
$$
\gamma(t) = (t, t)
$$
Suppose that there were a scaling function $\sigma$ on the plane that made $\gamma$ into a geodesic.
Since $\gamma'$ is constant, the first term in 
$$
0=\nabla_{\gamma'}\gamma'+ 2 (\gamma') \cdot \gamma'(\sigma) - g(\gamma', \gamma') \cdot \operatorname{grad} \sigma
$$
is zero [NB: Per the comments, this is in doubt]. The second term is just twice the directional derivative of $\sigma$ in the direction $[1, 1]^t$, times the vector $[1, 1]^t$. The $g$ part of the third term is $1 + t^2$, and the gradient is ... well, the gradient. 
But this gives us something of the form 
$$
c \begin{bmatrix}1\\1 \end{bmatrix} = (1+t^2) \operatorname{grad} \sigma
$$
which tells us that the two components of the gradient must be equal. So $\sigma$ is constant on the lines where $x + y = c$, hence we can write $$
\sigma(x, y) = f(x + y)
$$
for some function $f$. Now 
$$
\operatorname{grad} \sigma (x, y) = \begin{bmatrix} f'(x+ y) \\ f'(x+ y)\end{bmatrix}
$$
and so our equation becomes
\begin{align}
2 (\gamma' \cdot \sigma) \begin{bmatrix}1\\1 \end{bmatrix} &= (1+t^2) \operatorname{grad} \sigma \\
2 (2f'(t+ t)) \begin{bmatrix}1\\1 \end{bmatrix} &= (1+t^2) \begin{bmatrix} f'(t+ t) \\ f'(t+ t)\end{bmatrix}
\end{align}
That seems to say that $4 = 1 + t^2$ (or that $f$ is constant). The first is false, so $f$ must be constant ... but a constant $f$ does not make this path a geodesic. 
I've probably screwed up somewhere obvious, but ... that's what I've got. 
Post-comment addition
As mentioned, the claim about the directional derivative of $\gamma'$ is in doubt...but the rest of the analysis shows that this directional derivative, if it's not zero, must be in the $[1, 1]$ direction. 
So the question becomes: if you're walking home, and slogging through the mud in a non-optimal way, and someone alters the muddiness in a way that's constant along lines orthogonal to your path, can your route become optimal? I don't see how it could, but that's not a proof that it cannot. 
