It might be worthwhile to supplement Kevin Buzzard's answer by describing exactly what goes wrong in the case of $Q$ and $D_4$. Let $V$ and $W$ be the two dimensional irreducible representations of $Q$ and $D_4$ respectively. 

The representation $V \otimes V$ is isomorphic to $1 \oplus \chi_1 \oplus \chi_2 \oplus \chi_3$, where the $\chi_i$ are the nontrivial (one-dimensional) characters of $Q$. Similarly, $W \otimes W$ is isomorphic to $1 \oplus \eta_1 \oplus \eta_2 \oplus \eta_3$ where the $\eta_i$ are the nontrivial (one-dimensional) characters of $D_4$. The reason these formulas look the same is that the representation rings of $Q$ and $D_4$ are isomorphic.

However, we cannot build this isomorphism out of maps of vector spaces which commute with tensor product. Specifically, we cannot find an isomorphism $\alpha : V \to W$ of vector spaces so that the induced map $\alpha \otimes \alpha : V \otimes V \to W \otimes W$ carries $1$ to $1$ and $\chi_i$ to $\eta_i$. (Of course, I haven't told you how to label the $\chi_i$ and $\eta_i$. My statement is that there is no choice of labeling for which you can do this.)

This is very easy to see. The map $\alpha \otimes \alpha$ will carry the anti-symmetric elements of $V \otimes V$ to the anti-symmetric elements of $W \otimes W$. But $\wedge^2 V$  is the trivial representation, and $\wedge^2 W$ is not! So isomorphisms of the form $\alpha \otimes \alpha$ can't carry $1$ to $1$.