No. E.g., let $N=1$ and suppose that $X:=X_1$ has a nondegenerate zero-mean distribution $\mu$. Let $Y$ be an independent copy of $X$.

Then the expected $\mathcal W_p$-distance from the empirical distribution to $\mu$ is
$$E\mathcal W_p(\delta_X,\mu)^p=E|X-Y|^p>E|X|^p=E\mathcal W_p(\delta_X,\delta_0)^p;$$ the inequality here is an instance of a strict version of Jensen's inequality, which holds because the distribution $\mu$ is nondegenerate.

No. E.g., let $N=1$ and suppose that $X:=X_1$ has a nondegenerate zero-mean distribution $\mu$ such that $E|X|^p<\infty$. Let $Y$ be an independent copy of $X$.

Then the expected $\mathcal W_p$-distance from the empirical distribution to $\mu$ is
$$E\mathcal W_p(\delta_X,\mu)^p=E|X-Y|^p>E|X|^p=E\mathcal W_p(\delta_X,\delta_0)^p;$$ the inequality here is an instance of a strict version of Jensen's inequality, which holds because the distribution $\mu$ is nondegenerate.

No. E.g., let $N=1$ and suppose that $X:=X_1$ has a nondegenerate zero-mean distribution $\mu$. Let $Y$ be an independent copy of $X$.

Then for the expected $\mathcal W_p$-distance from the empirical distribution to $\mu$ one has
$$E\mathcal W_p(\delta_X,\mu)^p=E|X-Y|^p>E|X|^p=E\mathcal W_p(\delta_X,\delta_0)^p;$$ the inequality here is an instance of Jensen's inequality.

No. E.g., let $N=1$ and suppose that $X:=X_1$ has a nondegenerate zero-mean distribution $\mu$. Let $Y$ be an independent copy of $X$.

Then the expected $\mathcal W_p$-distance from the empirical distribution to $\mu$ is
$$E\mathcal W_p(\delta_X,\mu)^p=E|X-Y|^p>E|X|^p=E\mathcal W_p(\delta_X,\delta_0)^p;$$ the inequality here is an instance of a strict version of Jensen's inequality, which holds because the distribution $\mu$ is nondegenerate.