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A Lie derivative is applied to a function over a differentiable manifold. You can't have a symmetry of the function if you are not already working with a symmetry of the manifold itself. The Lie group of continuously generated symmetries of a differentiable manifold (with no structure other than topology) is the diffeomorphism group and the generators of diffeomorphism are the Lie derivatives. You identify a vector field with a Lie derivativ by treating the vectors as rates of point flows for the one parameter diffeomorphism subgroup generated by a Lie derivative.

SIDENOTE:I think there's still something unsettled here but only in so far as this defines vector and thence tensor fields

So if you are looking for a symmetry of a field you are looking for a subgroup of the diffeomorphism group which will map the field to itself, and this means the generators of that subgroup must, as infinitesimal actors, map the field to zero. For scalar fields:

[tex] \mathcal{D}_\epsilon \phi = \exp(\epsilon \mathcal{L}) \phi \approx ( \boldsymbol{1}+\epsilon \mathcal{L} ) \phi \to \phi[/tex]

for small [itex]\epsilon[/itex].

Now to transform a vector field you need a bit more... but by identifying vector fields with the Lie derivatives you have the natural adjoint action of the Lie derivative a.k.a. vector fields, on themselves in the form of the Lie (commutator) product. But the gist of it is that we can thereby extend the actions of Lie derivatives and thus diffeomorphisms to vector and thence tensor fields.

With this in mind here's the short answer to your question.

Given a tensor field [itex]\mathbf{A}[/itex] and a vector field [itex]X[/itex] with corresponding Lie derivative [itex]\mathcal{L}_X[/itex], the generated 1-parameter Lie subgroup: [itex]\{ \Gamma(\theta)=\exp(\theta\mathcal{L}_X)\}_{\theta \in \mathbb{R}}[/itex] is a symmetry of the field if each element of the group leaves the field invariant:

[tex] \Gamma(\theta)\mathbf{A} = \mathbf{A}\to[/tex]

which implies that the corresponding Lie derivative...

[tex]e^{\theta\mathcal{L}_X}\mathbf{A}=\left[\boldsymbol{1}+\theta \mathcal{L}_X + \theta^2 \mathcal{L}_X^2 +\ldots\right]\mathbf{A}

= \mathbf{A}+\boldsymbol{0}\to[/tex]

maps the field to the zero field.

[tex]\mathcal{L}_X \mathbf{A} = \boldsymbol{0}[/tex]

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