Using the singly and doubly ionized helium beams as well as a ${\mathrm{He}}^{3}$ beam from our electrostatic generator, we have studied the low-lying level structure of ${\mathrm{Na}}^{22}$ in the ${\mathrm{F}}^{19}(\ensuremath{\alpha}, n\ensuremath{\gamma}){\mathrm{Na}}^{22}$ and ${\mathrm{Ne}}^{20}({\mathrm{He}}^{3}, p\ensuremath{\gamma}){\mathrm{Na}}^{22}$ reactions, respectively. We found one excited state at 666\ifmmode\pm\else\textpm\fi{}4 kev in addition to most other states previously discovered in the self-conjugate reaction ${\mathrm{Mg}}^{24}(d, \ensuremath{\alpha}){\mathrm{Na}}^{22}$ which presumably yields only states having $T=0$. We believe this state to be the ${0}^{+}$ analog of the ground states of the two neighboring even-even nuclei forming the $T=1$ triplet at $A=22$. By various coincidence experiments, this state is found to decay by a 73-kev transition, having a probable half-life of 0.014 \ensuremath{\mu}sec, to the first-excited state of ${\mathrm{Na}}^{22}$ at 593 kev, which in turn decays to the ${3}^{+}$ ground state with a half-life of 0.266\ifmmode\pm\else\textpm\fi{}0.010 \ensuremath{\mu}sec. With the plausible assignment of ${1}^{+}$ for the 593-kev state (${0}^{+}$ being completely ruled out by the half-life) this would represent a pure $M1$---pure $E2$ cascade, whose strengths would be 0.0045 and 0.0077 single-particle units, respectively. The location of the first $T=1$ state occurs exactly as expected from a systematic study of the $A=4n+2$ series Coulomb-energy differences, lending additional support to the $I={0}^{+}$, $T=1$ assignment.We observed proton groups to most previously known states, and one new one at 3.75 Mev. A number of higher energy gamma rays were found in coincidence with those already mentioned which we can interpret within the framework of known excited states.