Suppose $L:K$ is a splitting field extension for a monic, separable polynomial $f \in K[t]$ which is irreducible over $K$. Let $\alpha_1, \cdots, \alpha_n$ be the distinct roots of $f$ in $L$. Let $k < n/2$ and choose a subset $\beta_1, \cdots, \beta_k, \delta_1, \cdots, \delta_k \subset \alpha_1, \cdots, \alpha_n$ of the roots (all the elements are distinct.)
I know that the maps in $Gal(L:K)$ permute the roots $\alpha_1, \cdots, \alpha_n$. Can I always find an element $\tau \in Gal(L:K)$ such that $\tau(\beta_1) = \delta_1, \cdots, \tau(\beta_k) = \delta_k$?
If not, how do I know which subsets of $\alpha_1, \cdots, \alpha_n$ can be mapped to which other subsets? (Is there a way of knowing this, in general?)
The short answer is no. Consider the cylotomic field extension $\mathbb{Q}(\zeta_p)$ over $\mathbb{Q}$ with Galois group $\mathbb{Z}_p^{\times}$, and basis $\zeta_p, ..., \zeta_p^{p-1}$. Then there is no automorphism that will send $\zeta_p$ to $\zeta_p^2$ and $\zeta_p^3$ to $\zeta_p^4$.
In general this depends on the structure of the Galois group. A permutation group on $n$ elements is called $k$-transitive if any ordered $k$-tuple can be mapped to any other ordered $k$-tuple (which is exactly what you have asked). Multiply transitive groups (specifically doubly transitive groups) are rare - all doubly transitive groups are known. There is a list of classifications here http://mathworld.wolfram.com/TransitiveGroup.html
There are extensions for which this is possible. For instance, Hilbert showed that $S_n$ and $A_n$ can represented as Galois groups of Galois extensions of $\mathbb{Q}$. (It is a nice exercise to check that $S_n$ is $n$-transitive and $A_n$ is $n-2$-transitive).