So knowing the $a_i$ which appear in the characteristic polynomial $\rho_V(\phi)$ completely determines the decomposition into irreducibles. The formula for the dimension of $Hom_G$ now follows from Schurs Lemma. (And I hope the same reasoning goes through for the continuous case as well)

If $V$ decomposes as a sum of irreducibles $V_i$, then the characteristic polynomial of $\rho_V(\phi)$ is the product of the characteristic polynomials $\rho_{V_i}(\phi)$. Since $G$ is abelian, all irreducible representations are 1-dimensional, therefore each irreducible $V_i$ gives you one eigenvalue $a_i$. Conversely, since $G$ is cyclic, the character of $V_i$ is completely defined by $a_i$ (the values are powers of it) hence $V_i$ is determined as well.

I have in fact just been brooding over your paper "Two-primary algebraic K-theory of pointed spaces" for a solution! I will not check this answer as accepted for now, try those calculations, and accept it in case they turn out to be fruitful.

This yields as the smallest example the group $Q_4$ with 16 elements, with $K_{−1}\mathbb Z Q_4 = \mathbb Z_2$. As a further remark, the point of this paper for the author was to calculate negative $K$-theory for finite subgroups of $SL_1(\mathbb H)$, so it would be interesting to understand how non-trivial torsion is linked to quarternionic representations.