Using the correspondences $K/Ad\ K \cong T/W \cong (\mathfrak t/Q^\vee)/W \cong \mathfrak t/(Q^\vee \rtimes W) = \mathfrak t/\hat W$, you can think about the Weyl alcove $A$ as parametrizing the conjugacy classes in the simply connected compact Lie group $K$.

$Z(K)$ obviously acts on the space of conjugacy classes, and since $Z(K) \leq T$ we can pick logarithms of $Z(K)$ inside $\mathfrak t$ to implement this action by translation on $\mathfrak t$, and therefore act on $A$ by rigid motions.

The vertices (and facets) of $A$ correspond to the nodes of the affine Dynkin diagram, so a rigid-motion action (preserving angles) gives an action on the diagram.

It's not obvious to me why (other than the classification) an affine Dynkin diaram automorphism is determined by where the "central" vertices go (those in the orbit of the affine vertex), but it's easily checked.

So now it's enough, given an element $z$ of $Z(K)$, to figure out which is the corresponding central vertex. I guess that amounts to picking a logarithm $X \in \mathfrak t$ with $\exp(X)=z$, and seeing where $X$ goes under the folding map $\mathfrak t \to A$. I'm a little afraid that this final step is exactly your question and that this may be an unsatisfying answer; maybe someone can provide a better one.

I believe the canonical reference for these ideas is Borel-de Siebenthal.

Using the correspondences $K/Ad\ K \cong T/W \cong (\mathfrak t/Q^\vee)/W \cong \mathfrak t/(Q^\vee \rtimes W) = \mathfrak t/\hat W =: A$, you can think about the Weyl alcove $A$ as parametrizing the conjugacy classes in the simply connected compact Lie group $K$. (Note that the penultimate equality only holds for $K$ simply connected.)

$Z(K)$ obviously acts on the space of conjugacy classes by multiplication, and since $Z(K) \leq T$ we can pick logarithms of $Z(K)$ inside $\mathfrak t$ to implement this action by translation on $\mathfrak t$, and therefore act on $A$ by rigid motions.

The vertices (and facets) of $A$ correspond to the nodes of the affine Dynkin diagram, so a rigid-motion action (hence angle-preserving) gives an action on the diagram.

It's not obvious to me why (other than the classification) an affine Dynkin diaram automorphism is determined by where the "central" vertices go (those in the orbit of the affine vertex), but it's easily checked. EDIT: As Jim Humphreys mentions, those vertices are the ones corresponding to minuscule fundamental representations.

So now it's enough, given an element $z$ of $Z(K)$, to figure out which is the corresponding central vertex of $A$. I guess that amounts to picking a logarithm $X \in \mathfrak t$ with $\exp(X)=z$, and seeing where $X$ goes under the folding-up map $\mathfrak t \to A$. I'm a little afraid that this final step is exactly your question and that this may be an unsatisfying answer; maybe someone can provide a better one.

I believe the canonical reference for these ideas is Borel-de Siebenthal.