\delta^\alpha_{\beta} is the identity tensor; its indices with respect to any basis are the identity matrix. If \Lambda^{\alpha}_{\enspace\beta} is a
Lorentz transformation, its inverse is represented in
Einstein's summation convention by
lowering one index and raising the other. (This convention for denoting the inverse applies only to Lorentz tarnsformations, not tensors in general.) With these notational rules
\delta^\alpha_\beta=\Lambda^\alpha_{\enspace\gamma}\,\Lambda_\beta^{\enspace\gamma}=\eta^{\nu\gamma}\eta_{\mu\beta}\,\Lambda^\alpha_{\enspace\gamma}\,\Lambda^{\mu}_{\enspace\nu}
where \eta_{\rho\sigma} is the
metric tensor for
Minkowski space, and has the same components as its inverse, \eta^{\rho\sigma}, namely
Diag(-1,1,1,1) or Diag(1,-1,-1,-1), depending on which
sign convention is used. The equation just says that the transformation composed with its inverse is (by definition) the identity tensor.
But
\delta'_{\alpha\beta}=\Lambda_{\alpha}^{\enspace \gamma}\,\delta_{\gamma\delta}\,\Lambda_{\beta}^{ \enspace \delta} \neq \delta_{\alpha\beta}
is rather the composition of the two matrices representing the Lorentz transformation. \delta_{\gamma\delta} is the identity matrix, but \delta'_{\alpha\beta} will not be, except in the trivial case where \Lambda_{\alpha}^{\enspace \gamma} is the identity matrix too.
