A famous theorem of Whitehead essentially states that spaces are determined by their homotopy groups. Is this true for spectra too?, i.e,
$$
\text{question: is a spectrum $E$ determined by its homotopy groups $\pi\_\*E$?}
$$



Are spectra determined by their homotopy groups?
Your statement of Whitehead's theorem is highly misleading. It is very easy to have inequivalent based spaces $X$ and $Y$ with $\pi\_\*(X)\simeq\pi\_\*(Y)$, and the same is true for spectra. For example, the spectra $KU$ and $\bigvee\_{n\in\mathbb{Z}}\Sigma^{2n}H\mathbb{Z}$ are inequivalent but have isomorphic homotopy (even as rings, if we give both spectra their usual ring structure). However, if $f\colon X\to Y$ is a map of spectra that induces an isomorphism $\pi\_\*(X)\to\pi\_\*(Y)$, then $f$ is an equivalence. The status of this statement depends somewhat on your framework of definitions. If we use orthogonal spectra with their usual model structure (or any similar framework), then it is a definition that $f$ is an equivalence if $\pi\_\*(f)$ is an isomorphism. It is a theorem that if $X$ and $Y$ are bifibrant and $f\colon X\to Y$ is an equivalence then there is a morphism $g\colon Y\to X$ and paths joining $g\circ f$ and $f\circ g$ to the respective identity maps in the relevant spaces of morphisms of orthogonal spectra. This is essentially the same as the case with spaces, where one needs to assume that $X$ and $Y$ have the homotopy type of a CW complex.


