Cotangent complex of perfect algebra over a perfect field

If $f: A \rightarrow B$ is a morphism of simplicial commutative rings (for example, a morphism of ordinary commutative rings), then the "topological" cotangent complex comes with additional structure: the action of $B$ on $L_{B/A}$ can be promoted to a (left) action of a certain associative ring spectrum $B^{+}$. There are canonical maps of ring spectra $B \rightarrow B^{+} \leftarrow \mathbf{Z}$ which induce (via the multiplication on $B^{+}$) an equivalence between $B^{+}$ and the smash product of $B$ with $\mathbf{Z}$ (beware that $B^{+}$ is not commutative and the order of the multiplication matters). There's also a canonical map of ring spectra $B^{+} \rightarrow B$, and the "algebraic" cotangent complex can be recovered as the tensor product $B \otimes_{ B^{+} } L_{B/A}$.

You can use this description (and the fact that $B^{+}$ is not too different from $B$) to answer a lot of the sorts of questions that you're asking. For example, $L_{B/A}$ is zero, or almost perfect, or connected through some range, if and only if the algebraic cotangent complex $L^{\mathrm{alg}}_{B/A}$ has the same property.

Let me explain why the $E_\infty$-cotangent complex $L_{B/A}$ vanishes for any map $A \to B$ of perfect rings over $\mathbf{F}_p$. (I do not know the answer to the more general question at the end.)

The proof uses formal properties of the cotangent complex (Kunneth formula, transitivity triangle) and relies on the following two observations (where all tensor products are derived):

1) If $R \to S$ is map of $E_\infty$-rings with $S \otimes_R S \simeq S$ via the multiplication map, then $L_{S/R} \simeq 0$. Indeed, we always have $L_{S \otimes_R S/R} \simeq p_1^* L_{S/R} \oplus p_2^* L_{S/R}$ by the Kunneth formula. If the multiplication map is an isomorphism, then we get $L_{S/R} \oplus L_{S/R} \simeq L_{S/R}$ via the sum map, which means $L_{S/R} \simeq 0$. (This is the classical proof that the cotangent complex of an open immersion is $0$.)

2) If $R \to S$ is any map of perfect rings, then $\pi_i(S \otimes_R S) =: \mathrm{Tor}^i_R(S,S)$ vanishes for $i > 0$. See, for example, Lemma 3.16 in https://arxiv.org/abs/1507.06490.

Now say $A \to B$ is a map of perfect rings. Consider the multiplication map $R := B \otimes_A B \to S := B$. Then $S \otimes_R S \simeq S$ via the multiplication map: this is clear on $\pi_0$ and thus follows from (2) as everything is perfect. Then (1) implies that $L_{B/B \otimes_A B} \simeq 0$. But the Kunneth formula and the transitivity triangle for $A \to B \otimes_A B \to B$ show that $L_{B/B \otimes_A B} \simeq L_{B/A}[1]$, and thus $L_{B/A} \simeq 0$.