Kronecker delta is not a tensor of any kind. It is just a fixed numerical matrix.
Tensors of order $#n#$ are linear functions from sets of $#n#$ vectors to the field of numbers and can be described in terms of any particular basis by arrays of numbers called components of the tensor with respect to that basis.
If the vector space has an inner product, then tensors of second order (ie ones involving two vectors) can be identified with linear operators on the vector space by requiring the matrix of the operator in basis $#\{ b_{i}\}#$ to be given by $#T_{ij}=\langle b_{i}|Tb_{j}\rangle=T(b_{i},b_{j})#$ .
In general, these components depend on the choice of basis, and the transformation rules for tensors tell us how they vary with changes in the basis that is being used to determine them.
For the special case where the tensor function $#T#$ is just the inner product, then for any orthonormal basis we get $#T_{ij}=T(b_{i},b_{j})=\langle b_{i}|b_{j}\rangle=\delta_{ij}#$, (and the corresponding operator is just the entity operator $#Tv=v#$ for all $#v#$). These components will be the same for any orthonormal basis, so the components are invariant under orthogonal transformations; but the same tensor may have a different matrix with respect to a basis that is either skew or not normalized. So what is true is that for a tensor whose components are given by the Kronecker delta, those components are invariant with respect to orthonormal changes of basis. (But, while the components will be changed by changes of scale or angle so that the new ones will not be given by$# \delta_{ij}#$, the Kronecker delta will still be what it was – just corresponding now to a different tensor.)