For a discrete group $\Gamma$, $T\in \mathbb{B}(l^{2}(\Gamma))$ is constant down the diagonals-meaning that for every $s, t, x, y\in \Gamma$ such that $ts^{-1}=yx^{-1}$, we have $\langle T\delta_{s}, \delta_{t} \rangle=\langle T\delta_{x}, \delta_{y} \rangle$.
Can we prove that: no nonzero compact operator in $\mathbb{B}(l^{2}(\Gamma))$ can be constant down the diagonals.
As Daniel mentioned, we have to assume that $\Gamma$ is infinite. Also, that the constant diagonal is not zero.
There is probably a more elegant way, but here is an argument. If you have such a compact $T$, you can assume that the constant diagonal is the main diagonal (otherwise you replace $T$ with $ts^{-1}T$ which is again compact, and $\langle ts^{-1}T\delta_s,\delta_t\rangle=\langle T\delta_s,st^{-1}\delta_t\rangle=\langle T\delta_s,\delta_s\rangle$).
As $T$ is compact, it is a norm-limit of finite-rank operators. This means that we can find a finite-rank operator $T_0$ such that all of its main diagonal entries are far from zero. But this is a contradiction, because a finite-rank operator is trace-class, and so the trace should be defined, which is not the case for $T_0$.