We have looked for the $J^P=0^{+}$ nuclear excitation of ${\mathrm{Zn}}^{68}$ in muonic atoms as suggested by Henley and Wilets. ${\mathrm{Zn}}^{68}$ was chosen because the energy of the $0^{+}$ excitation is close to the energy difference of the $2s-1s$ muonic levels, thus allowing a dynamic interaction between a muon in a $2s$ state and the nucleus in the ground state. We find that the $2s-1s$ level difference is 20 keV less than the nuclear excitation energy and, as a result, the dynamic interaction is too small for us to observe. We measure the ${\mathrm{Zn}}^{66}$-${\mathrm{Zn}}^{68}$ isotope shift to be $\frac{\delta R}{\delta R_{\mathrm{std}}}=0.47\mathrm{}\pm 0.11$.

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