If $X$ is co-H, then $\pi_1(X)$ must be a free group.  If $X$ is H, then $\pi_1(X)$
must be abelian.  The only free group that is abelian is $\mathbb{Z}$.

Argument for the first assertion:
The co-H structure (and Van Kampen) gives a factorization 
$$
\pi_1(X)\xrightarrow{i_*} \pi_1(X)*\pi_1(X)\xrightarrow{j_*} \pi_1(X)\times\pi_1(X)
$$
of the  map $\Delta_*$ induced by the diagonal $\Delta:X\to X\times X$.  This shows that $\pi_1(X)$ is isomorphic to a subgroup
of $G = (j_*)^{-1}( \mathrm{im}(\Delta_{*}))$, which is free on the elements 
$\{ x \bar x\}$  ($x$ and $\bar x$ represent the same element $x\in\pi_1(X)$ 
in the two summands of
$\pi_1(X)*\pi_1(X)$).  Now we are done because a subgroup of a free group is free.