Let $F$ be a field that we want to compute its rational algebraic $K$-theory using the Quillen's $Q$-construction. Let $QF$ be the $Q$ construction of the category of finite dimensional vector spaces over $F$. There is a rank filtration on $QF$. Because of a theorem of Solomon-Tits that says the homotopy type of the Tits building of a vector space is bouquet of spheres, this rank filtration gives the following spectral sequence for computing the rational homology of $QF$:

$E^1_{p,q}= H_q(GL_p(F^p),st(F^p))\otimes \mathbb{Q}=H_q((F^{\times})^p,\mathbb{Q})\implies H_{p+q}(QF)\otimes \mathbb{Q}$

For details of the spectral sequence you can see Theorem 4.3.3 here. It is kind of saying that the rational homology of $QF$ which contains the rational homotopy groups as primitive elements is determined by only multiplicative group of $F$.

Any integer $n$ acts on these homologies by raising the elements of $F^{\times}$ to the power of $n$. Raising to the power of $n$ acts on $H_q((F^{\times})^p,\mathbb{Q})$ by multiplication by $n^q$. So for each $n$ you can split the $H_i(QF,\mathbb{Q})$ into subspace that eigenvalue of the action of $n$ on those subspaces are $1,n,n^2,\cdots , n^{i-1}$ so subsequently these integers $n$ are acting on the rational algebraic $K$-theory such that $K_{i-1}(F)\otimes \mathbb{Q}$ splits to eigenspaces corresponding to the eigenvalues $1,n,n^2,\cdots , n^{i-1}$. As you can see these eigenspaces do not depend on the choice of $n$.

Does the action of $n$ defined above coincide with the action of the Adams operator $\psi^n$? Since they satisfy very similar properties but I cannot see how they are the same.