Let $U$ be an open subset of $\mathbb R^n$, and let $X$ be a vector field on $U$. You can construct a sheaf $\mathcal F$ of solutions of the ODE $Xf=0$ by letting $\mathcal F(U)$, for each open subset $V\subseteq U$, be the vector space of all $C^\infty$ functions $f$ on $V$ such that $Xf=0$.

By changing the field $X$ you can certainly change the isomorphism clas of $\mathcal F$.

Let $U=\mathbb R^2\setminus\{(0,0)\}$, define fields $X_1(x,y)=\Bigl((\frac12r-1)\frac xr-y,(\frac12r-1)\frac yr+x\Bigr)$ and $X_2(x,y)=(y,-x)$ and consider the corresponding sheaves $\mathcal F_1$ and $\mathcal F_2$. It is not difficult show show that $\mathcal F_1(U)$ is one-dimensional as a real vector space, while $\mathcal F_2(U)$ is infinite dimensional. It follows that $\mathcal F_1\not\cong\mathcal F_2$.

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Notice that $\mathcal F_1$ and $\mathcal F_2$ are locally isomorphic. This follows easily from the fact that the fields $X_1$ and $X_2$ are non-zero on their domain.

  [1]: http://img109.imageshack.us/img109/8818/plots.png