So, there are two distinct things that we're talking about here. It looks like you don't mind some advanced notation so I'll try to use that to illustrate the mathematical side of the physics I'm talking about.
Rigidity and the axis of rotation
One of the things that we're talking about is that the object is rigid, meaning that it's composed of a bunch of particles whose distances are fixed. The mathematical way to say tis is that its at any time must be represented by an isometry, and the group of isometries is $T(3) \times O(3).$ In fact, the isometry needs to be continuous with the identity so in normal infinite Euclidean space we can specify to $T(3) \times SO(3),$ translations plus rotations. Just to work out the basic idea, let Greek indices be coordinates and Latin indices be particles, so that we can use Einstein summation on Greek indices. I will try to use some explicit metric tensors $g_{\alpha\beta}$ to denote the dot product, to keep Latin and Greek indices visibly separate. The particles have position vectors $r_n^\alpha$ but the distances between those particles are constant, so $$\frac{d}{dt} \left[g_{\alpha\beta} ~(r_m^\alpha - r_n^\alpha)~(r_m^\beta - r_n^\beta)\right] = 2 g_{\alpha\beta} ~(r_m^\alpha - r_n^\alpha)~(\dot r_m^\beta - \dot r_n^\beta) = 0.$$Given an orientation (totally antisymmetric $[0,\;n]$ tensor $\epsilon$ on our $\mathbb R^n$ space) we can describe the rotation with a $[n-2,\;0]$ tensor $\Omega$ as$$\dot r^\beta_m = \chi^\beta + g^{\beta\gamma}\epsilon_{\gamma\delta\Lambda}~\Omega^\Lambda~r_m^\delta.$$This functional form causes the above term to be $\epsilon_{\alpha\delta\Lambda}~\Omega^\Lambda~R_{mn}^\alpha~R_{mn}^\delta = 0$ due to the antisymmetry of the $\epsilon$ term, where the exact form of $R^\alpha_{mn} = r^\alpha_m - r^\alpha_n$ doesn't matter, just as $\Omega^\Lambda$ doesn't matter, in deriving that $0$. It's purely from antisymmetry. In 2D, $\Omega^\Lambda$ is a scalar; in 3D it is a vector; in higher dimensions it is a tensor, but in each case it turns into this antisymmetric matrix $\Omega_{\gamma\delta} = \epsilon_{\gamma\delta\Lambda} ~\Omega^\Lambda.$ The fact that the rotations can all be represented by these antisymmetric matrices is going to be very useful in a moment. I am not sure whether in higher dimensions other terms also pop out; my thinking was just "you either need velocity to vanish directly or to be perpendicular to the position."
Angular momentum
Another is, due to the fact that the formula for kinetic and potential energies do not depend on rotation (a continuous symmetry), there is a conserved Noether current associated with that symmetry: angular momentum. In this case our Lagrangian has the form $$\frac12 \sum_n m_n g_{\alpha\beta} \dot r_n^\alpha \dot r_n^\beta - \sum_{mn} U_{mn}\left[g_{\alpha\beta} ~ (r_m^\alpha - r_n^\alpha)~(r_m^\beta - r_n^\beta)\right],$$
for some set of strong potentials which force the body to remain rigid $U_{mn}.$ Since those are rotationally symmetric, and the kinetic term is rotationally symmetric, Noether's theorem says that we therefore pick up a conserved quantity for any symmetry-obeying infinitesimal displacement $\delta r^\alpha$ of $$Q = \sum_n \frac{\partial L}{\partial \dot r_n^\alpha} ~\delta r_n^\alpha.$$Again, the rotation turns no matter what into an antisymmetric matrix, $\delta r_n^\alpha = g^{\alpha\mu}~\delta\phi_{\mu\nu}~r_n^{\nu},$ so the above becomes:$$Q = \sum_n m_n g_{\alpha\beta}~\dot r_n^\beta ~g^{\alpha\mu}~\delta\phi_{\mu\nu}~r_n^{\nu}= \delta\phi_{\mu\nu}~\sum_n m_n~\dot r_n^\mu ~r_n^{\nu} = \delta\phi_{\mu\nu}~Q^{\mu\nu}.$$
This conserved quantity therefore has a $[2,\;0]$ tensor character, as we can choose any axis for this rotation. Moreover, any symmetric part to this tensor will get nuked by the antisymmetry of the rotation, so without loss of generality we can antisymmetrize it, too. The usual notation for this is to write $Q^{[\mu\nu]}$ with square brackets, $$Q^{[\mu\nu]} = \frac12 (Q^{\mu\nu}-Q^{\nu\mu})=\sum_n m_n ~\dot r_n^{[\mu} ~r_n^{\nu]}.$$
Bringing them together
We've seen two fundamentally different expressions here: one is the angular velocity tensor $\Omega_{\alpha\beta}$, which stems from the rigidity of the system; the other is the angular momentum tensor $Q^{\mu\nu},$ which stems from the symmetry in the equations of motion. They're obviously not defined the same, but they both turn out to be antisymmetric. What is the relation between them? That's easy: substitute the rotational term from the $\Omega$ expression for $\dot r_n^\beta$ into the same $\dot r$ term in the kinetic energy to find:$$Q^{[\mu\nu]} = -\left[\sum_n m_n~r_n^{[\nu}~g^{\mu]\gamma}~r_n^{\delta} \right]~\Omega_{\gamma\delta}.$$Here we see that a $[4,\;0]$ tensor is linearly relating them and it has $\nu$-$\delta$ and $\mu$-$\gamma$ symmetry but $\nu$-$\mu$ antisymmetry. So they always stand in a direct relationship via this moment-of-inertia tensor, and that's basically because particle velocities about the center of mass both contribute directly to the angular momentum and are directly determined by a rotation.
Of course in $\mathbb R^3$ it happens to be easier to work with the angular momentum and angular velocity vectors; we write $Q^{\mu\nu}$ as $Q_\lambda~\epsilon^{\lambda\mu\nu}$ and notice that for the most common orientation (where $\epsilon_{123} = 1$) we have $\epsilon^{\alpha\beta\gamma}\epsilon_{\beta\gamma\delta} = 2\delta^\alpha_\delta,$ so that $$Q_\lambda = \frac12 \left[\sum_n m_n~\epsilon_{\lambda\mu\nu}~g^{\mu\gamma}~r_n^\nu ~r_n^\delta~\epsilon_{\gamma\delta\kappa} \right] \Omega^\kappa.$$So in $\mathbb R^3$ we find a direct $[0,\;2]$-tensorial relationship between the same quantities, because both of the angular-momentum matrices are secretly angular-momentum vectors.
I'm not entirely sure I've done all of the details properly here, but that's the general story. The two concepts are different, in part because one contains "mass" ideas that are different in different directions and the other does not; but they turn out to be linearly related through the terms $\dot r_n^\alpha$. They are secretly antisymmetric matrices with a linear relation, but they can be down-converted in $\mathbb R^2$ to scalars or in $\mathbb R^3$ to vectors or in $\mathbb R^4$ to pairs of vectors in $\mathbb R^3.$
