Let $\tilde A(X)$ be the reduced functor, i.e., the homotopy fiber of the map $A(X) \to A(\ast)$. Since $A(*)$ is rationally a product of $K(Q,4j+1)$ for $j \ge 1$, we may as well study $\tilde A(X)$ instead.

The rational homotopy of $\Omega \tilde A(\Sigma Y)$ was studied in

The free loop space and the algebraic K-theory of spaces. K-Theory 1 (1987), no. 1, 53–82.

(The paper has some gaps but these were later corrected.) If we assume that $Y$ is connected, then the rational homotopy type of $\Omega \tilde A(\Sigma Y)$ coincides with that of the functor: $$ \prod_{n\ge 1} Q(Y^{[n]}_{h\Bbb Z_n}) $$ where $Y^{[n]}$ is the $n$-fold smash product of $Y$, $\Bbb Z_n$ acts by cyclic permutation and ${({-})}_{h\Bbb Z_n}$ means homotopy orbits. Rationally, the homotopy groups of $Q(Y^{[n]}_{h\Bbb Z_n})$ coincide the homology groups of $Y^{[n]}_{\Bbb Z_n}$. When $Y$ is an $j$-sphere, the rational homology is not hard to compute.

Finally, if $X = S^1$, we know that $\tilde A(S^1)$ is rationally the same as $B A(*)$ by a version of Bass-Heller-Swan. But by what I mentioned above, $B\tilde A(*)$ is a rationally the product of $K(\Bbb Q,4j+2)$, $j \ge 1$.

Let $\tilde A(X)$ be the reduced functor, i.e., the homotopy fiber of the map $A(X) \to A(\ast)$. Since $A(*)$ is rationally a product of $K(Q,4j+1)$ for $j \ge 1$, we may as well study $\tilde A(X)$ instead.

The rational homotopy of $\Omega \tilde A(\Sigma Y)$ was studied in

G. Carlsson, R. Cohen, T. Goodwillie, and W. Hsiang, The free loop space and the algebraic K-theory of spaces. K-Theory 1 (1987), no. 1, 53–82.

(The paper has some gaps but these were later corrected.) If we assume that $Y$ is connected, then the rational homotopy type of $\Omega \tilde A(\Sigma Y)$ coincides with that of the functor: $$ \prod_{n\ge 1} Q(Y^{[n]}_{h\Bbb Z_n}) $$ where $Y^{[n]}$ is the $n$-fold smash product of $Y$, $\Bbb Z_n$ acts by cyclic permutation and ${({-})}_{h\Bbb Z_n}$ means homotopy orbits. Rationally, the homotopy groups of $Q(Y^{[n]}_{h\Bbb Z_n})$ coincide the homology groups of $Y^{[n]}_{\Bbb Z_n}$. When $Y$ is a $j$-sphere, the rational homology is not hard to compute.

Finally, if $X = S^1$, we know that $\tilde A(S^1)$ is rationally the same as $B A(*)$ by a version of Bass-Heller-Swan. But by what I mentioned above, $B\tilde A(*)$ is a rationally the product of $K(\Bbb Q,4j+2)$, $j \ge 1$.

Let $\tilde A(X)$ be the reduced functor, i.e., the homotopy fiber of the map $A(X) \to A(\ast)$. Since $A(*)$ is rationally a product of $K(Q,4j+1)$ for $j \ge 1$, we may as well study $\tilde A(X)$ instead.

The rational homotopy of $\Omega \tilde A(\Sigma Y)$ was studied in

The free loop space and the algebraic K-theory of spaces. K-Theory 1 (1987), no. 1, 53–82.

(The paper has some gaps but these were later corrected.) If we assume that $Y$ is connected, then the rational homotopy type of $\Omega \tilde A(\Sigma Y)$ coincides with that of the functor: $$ \prod_{n\ge 1} Q(Y^{[n]}_{h\Bbb Z_n}) $$ where $Y^{[n]}$ is the $n$-fold smash product of $Y$, $\Bbb Z_n$ acts by cyclic permutation and ${}_{h\Bbb Z_n}$ means homotopy orbits. Rationally, the homotopy groups of $Q(Y^{[n]}_{h\Bbb Z_n})$ coincide the homology groups of $Y^{[n]}_{\Bbb Z_n}$. When $Y$ is an $j$-sphere, the rational homology is not hard to compute.

Finally, if $X = S^1$, we know that $\tilde A(S^1)$ is rationally the same as $B A(*)$ by a version Bass-Heller-Swan. But by what I mentioned above, $B\tilde A(*)$ is a rationally the product of $K(\Bbb Q,4j+2)$, $j \ge 1$.

Let $\tilde A(X)$ be the reduced functor, i.e., the homotopy fiber of the map $A(X) \to A(\ast)$. Since $A(*)$ is rationally a product of $K(Q,4j+1)$ for $j \ge 1$, we may as well study $\tilde A(X)$ instead.

The rational homotopy of $\Omega \tilde A(\Sigma Y)$ was studied in

The free loop space and the algebraic K-theory of spaces. K-Theory 1 (1987), no. 1, 53–82.

(The paper has some gaps but these were later corrected.) If we assume that $Y$ is connected, then the rational homotopy type of $\Omega \tilde A(\Sigma Y)$ coincides with that of the functor: $$ \prod_{n\ge 1} Q(Y^{[n]}_{h\Bbb Z_n}) $$ where $Y^{[n]}$ is the $n$-fold smash product of $Y$, $\Bbb Z_n$ acts by cyclic permutation and ${({-})}_{h\Bbb Z_n}$ means homotopy orbits. Rationally, the homotopy groups of $Q(Y^{[n]}_{h\Bbb Z_n})$ coincide the homology groups of $Y^{[n]}_{\Bbb Z_n}$. When $Y$ is an $j$-sphere, the rational homology is not hard to compute.

Finally, if $X = S^1$, we know that $\tilde A(S^1)$ is rationally the same as $B A(*)$ by a version of Bass-Heller-Swan. But by what I mentioned above, $B\tilde A(*)$ is a rationally the product of $K(\Bbb Q,4j+2)$, $j \ge 1$.