Lorentz transformations are linear operators on ##\mathbb R^4## (or ##\mathbb R^2## or ##\mathbb R^3##). So they can be represented by matrices. (See the https://www.physicsforums.com/showthread.php?t=694922 about matrix representations of linear transformations). I will not make any notational distinction between a linear operator and its matrix representation with respect to the standard basis.
Let ##\Lambda## be an arbitrary Lorentz transformation. By definition of Lorentz transformation, we have ##\Lambda^T\eta\Lambda=\eta##. This implies that ##\Lambda^{-1}=\eta^{-1}\Lambda^T\eta##. Let's use the notational convention that for all matrices X, we denote the entry on row ##\mu##, column ##\nu## by ##X^\mu{}_\nu##. If we use this convention, the definition of matrix multiplication, our formula for ##\Lambda^{-1}## and the convention that every index that appears twice is summed over, we get
$$(\Lambda^{-1})^\mu{}_\nu = (\eta^{-1})^\mu{}_\rho (\Lambda^T)^\rho{}_\sigma \eta^\sigma{}_\nu = (\eta^{-1})^\mu{}_\rho \Lambda^\sigma{}_\rho \eta^\sigma{}_\nu.$$ This is where things get funny. It's conventional to write ##\eta_{\mu\nu}## instead of ##\eta^\mu{}_\nu##, and ##\eta^{\mu\nu}## instead of ##(\eta^{-1})^\mu{}_\nu##. If we use this convention, we have
$$(\Lambda^{-1})^\mu{}_\nu = \eta^{\mu\rho} \Lambda^\sigma{}_\rho \eta_{\sigma\nu}.$$ Now if we also use the convention that ##\eta^{\mu\nu}## raises indices and ##\eta_{\mu\nu}## lowers them, we end up with
$$(\Lambda^{-1})^\mu{}_\nu = \Lambda_\nu{}^\mu.$$ So if ##\Lambda## isn't the identity transformation, we have
$$\Lambda_\nu{}^\mu = (\Lambda^{-1})^\mu{}_\nu \neq \Lambda^\mu{}_\nu.$$ As you can see, the inequality is a result of the definitions of ##\eta_{\mu\nu}## and ##\eta^{\mu\nu}##, so if you use a notational convention that denotes these things by something else, or doesn't use these things to raise and lower indices, it may be OK to write ##\Lambda^\mu_\nu##.