Does the Green's function of the simple random walk on $\mathbb Z^d$ always vary locally?

We will use the elementary fact that for $m \ge k \ge m/2$, the binomial coefficients satisfy $${m\choose k+1} <{m \choose k}. \quad (\#)$$

The case $x=0$ is obvious so we may assume $x$ has some nonzero coordinate. By symmetry, we may assume that $x_1>0$. Then it suffices to show that for every $y \in \mathbb Z^d$ that satisfies $y_1 \ge 0$, the point $z=z(y)$ that agrees with $y$ in all coordinates except the first, where $z_1=y_1+2$, we have the strict inequality $$P(S_n=z)<P(S_n=y)$$ for all $n$ such that $P(S_n=z)>0$. Let $A_n$ be the (random) set of steps among the first $n$ when the random walk moved in the first coordinate, and let $w^*$ denote the projection of a node $w \in\mathbb Z^d$ to coordinates $2,3 ,\ldots, d$. Fix $y$ with $y_1 \ge 0$ and let $z=z(y)$ as above, so that $z^*=y^*$. If $A_n$ satisfies $$P(S_n=z \,|A_n)>0 \,,$$ then the cardinality $|A_n|$ and $z_1$ must have the same parity, and by $(\#)$, $$P(S_n=z\, | \, A_n)=P(S_n^*=z^* \,|A_n) \cdot {|A_n| \choose \frac{|A_n|+z_1}{2}}2^{-|A_n|} \; < $$ $$ \: P(S_n^*=y^* \,|A_n) \cdot {|A_n| \choose \frac{|A_n|+y_1}{2}}2^{-|A_n|}= P(S_n=y\, | \, A_n) \,.$$ Taking expectations (i.e., averaging over $A_n$) gives $P(S_n=z)<P(S_n=y)$.