If we let $S: TM \to TM$ denote the shape operator of a hypersurface $M \to \mathbb R^{n+1}$ and $\mathfrak R:\Lambda^2TM\to\Lambda^2TM$ the curvature operator, then the Gauss equation can be written $\mathfrak{R} = \Lambda^2 S.$ Here $\Lambda^2S$ is the obvious induced map on bivectors $(\Lambda^2 S)(v \wedge w) = Sv \wedge Sw.$

Thus the (pointwise) intrinsic Riemannian invariants of $S$ are exactly those that can be written in terms of $\Lambda^2 S$.

Since $S$ is self-adjoint, we can choose an orthonormal basis $e_i$ for $T_p M$ so that $S = \mathrm{diag}(\kappa_1,\ldots,\kappa_n)$. Just as we can get the standard mean curvature by taking the trace of $S$, we can get the higher invariants of $S$ by taking traces of $\Lambda^k S \colon$ we compute

$$\begin{align} \mathrm{tr}(\Lambda^k S) &= \sum_{i_1<\ldots<i_k} \langle e_{i_1}\wedge\cdots\wedge e_{i_k}, (\Lambda^k S)(e_{i_1}\wedge\cdots\wedge e_{i_k}) \rangle \\ &= \sum_{i_1<\ldots<i_k}\langle e_{i_1}, S e_{i_1} \rangle\cdots\langle e_{i_k}, S e_{i_k} \rangle\\ &= \sum_{i_1<\ldots<i_k}\kappa_{i_1}\cdots \kappa_{i_k} \end{align}$$ which is (perhaps up to a constant) the "higher order mean curvature" $H_k$. Thus we can confirm your conjecture that this is intrinsic for even $k$: in this case we can write $$(\Lambda^k S)(v_1\wedge\ldots\wedge v_k) = (\Lambda^2 S)(v_1 \wedge v_2)\wedge \cdots \wedge(\Lambda^2 S)(v_{k-1}\wedge v_k),$$ so $\Lambda^2 S$ determines all the even invariants. On the other hand, when $k$ is odd, the transformation $S \to -S$ leaves $\Lambda^2 S$ invariant but changes the sign of $\Lambda^k S$, so the odd invariants of $S$ are not intrinsic Riemannian invariants.

This is perhaps a bit of a cop-out, since we can change the sign of $S$ just by flipping our unit normal while leaving all the geometry the same. Perhaps we should really ask whether $|H_k|$ is intrinsic  - this is true, for example, when $k=n$ is odd, in which case this is the Gauss-Kronecker curvature. I don't know the general answer to this question offhand.

If we let $S: TM \to TM$ denote the shape operator of a hypersurface $M \to \mathbb R^{n+1}$ and $\mathfrak R:\Lambda^2TM\to\Lambda^2TM$ the curvature operator, then the Gauss equation can be written $\mathfrak{R} = \Lambda^2 S.$ Here $\Lambda^2S$ is the obvious induced map on bivectors $(\Lambda^2 S)(v \wedge w) = Sv \wedge Sw.$

Thus the (pointwise) intrinsic Riemannian invariants of $S$ are exactly those that can be written in terms of $\Lambda^2 S$.

Since $S$ is self-adjoint, we can choose an orthonormal basis $e_i$ for $T_p M$ so that $S = \mathrm{diag}(\kappa_1,\ldots,\kappa_n)$. Just as we can get the standard mean curvature by taking the trace of $S$, we can get the higher invariants of $S$ by taking traces of $\Lambda^k S \colon$ we compute

$$\begin{align} \mathrm{tr}(\Lambda^k S) &= \sum_{i_1<\ldots<i_k} \langle e_{i_1}\wedge\cdots\wedge e_{i_k}, (\Lambda^k S)(e_{i_1}\wedge\cdots\wedge e_{i_k}) \rangle \\ &= \sum_{i_1<\ldots<i_k}\langle e_{i_1}, S e_{i_1} \rangle\cdots\langle e_{i_k}, S e_{i_k} \rangle\\ &= \sum_{i_1<\ldots<i_k}\kappa_{i_1}\cdots \kappa_{i_k} \end{align}$$ which is (perhaps up to a constant) the "higher order mean curvature" $H_k$. Thus we can confirm your conjecture that this is intrinsic for even $k$: in this case we can write $$(\Lambda^k S)(v_1\wedge\ldots\wedge v_k) = (\Lambda^2 S)(v_1 \wedge v_2)\wedge \cdots \wedge(\Lambda^2 S)(v_{k-1}\wedge v_k),$$ so $\Lambda^2 S$ determines all the even invariants. On the other hand, when $k$ is odd, the transformation $S \to -S$ leaves $\Lambda^2 S$ invariant but changes the sign of $\Lambda^k S$, so the odd invariants of $S$ are not invariant in this strict sense.

This is perhaps a bit of a cop-out, though, since we can change the sign of $S$ just by flipping our unit normal while leaving all the geometry the same. Thus, I think we should really ask whether $|H_k|$ is intrinsic. Since any rank-1 $S$ induces $\Lambda^2 S = 0$, this is clearly false for $k=1$; so let's assume $k \ge 3$. Noting that the unknown $a = \mathrm{rank}(S)$ and the known $b = \mathrm{rank}(\mathfrak R)$ are related by $b = \binom a 2$ (just look at their diagonalizations), we can handle two cases separately:

• If $b < 3$, then $\mathrm{rank}(S) < 3 \le k$ and thus $\Lambda^k S = 0$, so $|H_k| = 0$ is determined by $\mathfrak R.$
• If $b \ge 3$, then there's a nice rigidity theorem that tells us $\mathfrak R$ determines $S$ up to sign - see the first theorem in Chapter 12 of Spivak vol 5, or this MSE answer of mine.

Either way, we can determine $|H_k|$ from $\mathfrak R$; so all the higher odd invariants (excluding the actual mean curvature) are intrinsic up to their sign.