In such a reaction, the oxidation number are not modified. This dissolution is not a redox reaction. It will not produce $\ce{Cl2}$ and $\ce{UO2}$ (as stated by Young), which would imply an oxidation of Chlorine by Uranium. To find the equation, consider that autunite is made of the following ions :
$$\ce{Ca(UO2)2(PO4)2 -> Ca^{2+} + 2 UO2^{2+} + 2 PO4^{3-}}$$ When adding $\ce{HCl}$ solution, the $\ce{H+}$ ions will react with the phosphate ion and it is the main change. If a very small amount of $\ce{HCl}$ is added, this addition will produce a precipitate of $\ce{UO2HPO4}$, and not $\ce{CaHPO4}$ as proposed as proposed by Young (there is too many phosphate ions in autunite - the only calcium phosphate which can be formed here is $\ce{Ca(H2PO4)2}$ - see later on). The reaction is :
$$\ce{Ca(UO2)2(PO4)2 + 2 HCl -> CaCl2(aq) + 2 UO2HPO4(s)}$$
If the solution is concentrated enough, and with adding more $\ce{HCl}$, the following reaction may happen, remembering that $\ce{Ca(H2PO4)2}$ is only weakly soluble into water :
$$\ce{Ca(UO2)2(PO4)2 + 4 HCl -> Ca(H2PO4)2 (s) + 2 UO2Cl2(aq)}$$
But, as soon as some excess of $\ce{HCl}$ is added, both precipitates are redissolved according to the following global equation : $$\ce{Ca(UO2)2(PO4)2 + 6 HCl -> CaCl2(aq) + 2 UO2Cl2(aq) + 2H3PO4(aq)}$$
As the final substances are all soluble, they are dissociated into ions, so that the last equation should be better written, without $\ce{Cl-}$ ions :
$$\ce{Ca(UO2)2(PO4)2 + 6 H+ -> Ca^{2+}(aq) + 2 UO2^{2+}(aq) + 2H3PO4(aq)}$$
Ref.: F.-P. Treadwell, Analytical Chemistry, Qualitative Analysis, Dunod 1924, p. 142 - 145.
P. S. : Today it is difficult to find documents about the chemistry of Uranium compounds and their analytical properties.