Here is my example of length space with $pi_1(X)=mathbb Z_{3cdot 2^n}$ such that $mathop{diam}tilde X$ has order of $ncdot mathop{diam} X$ --- I can not do better.
(One can easely make a 4-dimensional manifold out of this.)
Consider sequence $tau_n$ of triangulations of disc, defined inductively on the following way:
Clearly the boundary of $tau_n$ consist of $3cdot 2^n$ edges.
Let us take cyclic sequence of $3cdot 2^n$ copies of $tau_n$ and glue each to the next one along the boundary, rotating by one edge.
On the obtained a 2-dimensional complex, change each triangle to a Reuleaux triangle of width 1, we obtaine space $tilde X$.
On $tilde X$, we have an free isometric action of $mathbb Z_{3cdot 2^n}$, it acts by shifting sequence of $tau_n$'s. Take $X=tilde X/mathbb Z_{3cdot 2^n}$. It straight forward to see that and $mathop{daim} Xle 2$ and $mathop{diam}tilde Xge tfrac n2$ (the distance from vertex $0$ to vertex $ell$ is the least number of terms in presetation of $ell$ as a sum of $pm 2^s$)...
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