When do the polynomial algebra and free algebra coincide in brave new algebra? Given an $\mathbb E_\infty$-ring (highly structured commutative ring spectrum if you want) $R$, we have the free $R$-algebra (on one generation) $R\{t\}\simeq \bigoplus_{n\ge 0} R_{\mathrm h\Sigma_n}$ vs. the polynomial $R$-algebra $R[t]\simeq \bigoplus_{n\ge 0} R.$
It is well-known that $R\{t\}\simeq R[t]$ when $R$ is rational, i.e. a $H\mathbb Q$-algebra. It is also very easy to verify that this is not the case in many other cases, for instance when $R= H\mathbb Z$ and $H\mathbb F_p$.
Q: Does $R\{t\}\simeq R[t]$ for a (connective?) $\mathbb E_\infty$-ring imply that $R$ is a $H\mathbb Q$-algebra?
In case not, I would be very happy with some example of non-rational ring spectra $R$ for which polynomial and free $R$-algebras agree. Also, in that case, what condition on an $\mathbb E_\infty$-ring $R$ does the condition that $R\{t\}\simeq R[t]$ imply?
In the special case of a discrete ring $R$ = Eilenberg-MacLane $\mathbb E_\infty$-ring $HR,$ this question reduces to one about group cohomology of symmetric groups. More precisely:
Q': Does the condition that $\mathrm H^i(\Sigma_n, R)\simeq 0$ for all $n$ and all $i\ge 1$ for a commutative ring $R$ imply that $R$ is a $\mathbb Q$-algebra?
This sounds like a possibly-very-easy piece of classical algebra, but to my shame, I do not know the answer even in this particular case.
 A: Any morphism of $R$-algebras $\varphi : R\{t\} \to R[t]$ is determined up to homotopy by an element of $\pi_0(R[t]) \approx \pi_0(R)[t]$.  If $\varphi$ is an equivalence, then this element must be the generator $t$, so we may as well assume $\varphi$ is the canonical map $\varepsilon_R : R\{t\} \to R[t]$.
Claim:
Let $R$ be an $E_\infty$-ring spectrum.  Then the map $\varepsilon_R$ is invertible if and only if $R$ is a $\mathbf{Q}$-algebra.
Useful observations:
1) The map $\varepsilon_R$ is compatible with extensions of scalars.  Therefore if $\varepsilon_R$ is invertible, then $\varepsilon_{R'}$ is invertible for any $R$-algebra $R'$.
2) $R$ is a $\mathbf{Q}$-algebra if and only if $\pi_0(R)$ is a $\mathbf{Q}$-algebra (see comments).
Sufficiency:
Suppose that $R$ is a $\mathbf{Q}$-algebra.  The map $\varepsilon_R$ is obtained by extension of scalars along the connective cover $R_{\ge 0} \to R$, so we may assume $R$ is connective.  The $\infty$-category of connective $E_\infty$-algebras over $\mathbf{Q}$ is equivalent to that of simplicial commutative $\mathbf{Q}$-algebras.  Under this equivalence, the map $\varepsilon_R$ corresponds to the identity morphism $R[t] \to R[t]$ (where by abuse of notation $R$ also denotes the corresponding simplicial commutative $\mathbf{Q}$-algebra).
Necessity:
Suppose $\varepsilon_R$ is an equivalence.  Since the map $\varepsilon_R$ is compatible with formation of connective covers, we may replace $R$ by its connective cover.  We may also extend scalars along $R \to \pi_0(R)$ to assume $R$ is discrete.
Consider any residue field $R \to k$.
If $k$ is of characteristic $p>0$, then $\varepsilon_k$ cannot be invertible.  Indeed, it is well-known that $\varepsilon_{\mathbf{F}_p}$ is not invertible, and $\varepsilon_k$ is the extension of scalars along the faithfully flat map $\mathbf{F}_p\{t\} \to k\{t\}$.
Thus every residue field $k$ must be of characteristic zero, so $R$ is a $\mathbf{Q}$-algebra.
