A vector field $X$ is a Killing field if the Lie derivative with respect to $X$ of the metric $g$ vanishes.
But if we're on an Euclidean space the metric tensor is constant, so it's Lie derivative on every vector field should vanish. If I work on local coordinates instead I find that only source-free vector fields are Killing.
Where am I doing something wrong?
The Euclidean metric in standard Euclidean coordinates is constant, but this does not mean it is ``Lie constant'' for every vector field $X$. Indeed, using the Leibniz rule for the Lie derivative of tensors, we get $$ X(g(Y, Z)) = (L_X g)(Y, Z) + g([X, Y], Z) + g(Y, [X, Z]). $$ If we let $Y = \partial_i$ and $Z = \partial_j$ be standard coordinate vector fields, then this becomes $$ Xg_{ij} = (L_X g)_{ij} + g([X, \partial_i], \partial_j) + g(\partial_i, [X, \partial_j]). $$ If $L_Xg = 0$ then this gives us that $g([X, \partial_i], \partial_j) + g(\partial_i, [X, \partial_j]) = Xg_{ij}$. In the case of the standard Euclidean metric in standard coordinates, we have $g_{ij} = 0$ and thus the right hand side vanishes. At this point we use that the Lie bracket in coordinates is $$ [X, Y] = (X^k\partial_k Y^\ell - Y^k \partial_k X^\ell)\partial_\ell $$ so $$ [X, \partial_i] = -\delta_i^k \partial_k X^\ell \partial_\ell = -\partial_i X^\ell \partial_\ell, \ \ \ \ [X, \partial_j] = -\partial_j X^\ell \partial_\ell. $$ Thus our equation above for $L_X g = 0$ becomes $$ \partial_i X^\ell g_{\ell j} + \partial_j X^\ell g_{\ell i} = 0. $$ But we know the metric components are $g_{ij} = \delta_{ij}$, so this is saying that $$ \partial_i X^j + \partial_j X^i = 0, $$ which is the condition $X$ must satisfy to be a Killing field. Indeed, Killing's equation in this case reads exactly this. Note that setting $i = j$ and summing gives $\text{div}X = 0$.