I will try to follow Poutnik's suggestion. And try to recognize the solutions without using external reagents. The only allowed mixtures are made by mixing two unknown solutions. Let's consider for the present article that the solutions are numbered $(1)$ to $(10)$, by alphabetic order. So the ten solutions are : $(1)$ = $\ce{AgNO3}$, $(2)$ = $\ce{Ag2SO4}$, $(3)$ = $\ce{Ba(OH)2}$, $(4)$ = $\ce{KCl}$, $(5)$ = $\ce{KI}$, $(6)$ = $\ce{NaOH}$, $(7)$ = $\ce{Na2SO4}$, $(8)$ = $\ce{NH4Cl}$ $(9)$ = $\ce{NH4I}$, $(10)$ = $\ce{Pb(NO3)2}$. But this correspondance is not known to the student. It has to be established by the student, who just knows the different substances, but not their order.
It should be mentioned that $\ce{Ag2SO4}$ is only poorly soluble in water : $0.57$ g in $100$ mL at $15°$C. This is about $0.0018$ M. The silver concentration in saturated $\ce{Ag2SO4}$ solutions (and in $(2)$) is then [$\ce{Ag+}$] = $0.0036$ M. Hopefully the other unknown solutions are more concentrated.
$45$ mixtures hcve to be done with these $10$ unknown solutions. A lot of these mixtures produce a precipitate. Let's look first at the mixtures producing a deep yellow precipitate (typical to $\ce{PbI2}$), which is produced twice : in the mixtures $(5)+(10)$, and $(9)+(10)$. As $(10)$ is used in both mixtures, $(10)$ must contain $\ce{Pb(NO3)2}$. Then $(5)$ and $(9)$ must contain $\ce{KI}$ or $\ce{NH4I}$. The only way of separating both iodides without adding chemicals is to heat and evaporate the pure solutions. If a solid remains in the test tube, it is ($5$), or $\ce{KI}$. If the final dry substance slowly sublimates, it is ($9$), or $\ce{NH4I}$. Now we know three substances : $\ce{Pb(NO3)2}$, $\ce{KI}$ and $\ce{NH4I}$.
Then, some mixtures produce a pale yellow precipitate ($\ce{AgI}$), which is observed in four mixtures : $(1)$ + $(5)$, $(1$) + $(9)$, $(2)$ + $(5)$ and $(2)$ + $(9)$. As we know that $(5)$ and $(9)$ contains the same anion iodide, $(1)$ and $(2)$ must contain $\ce{Ag+}$ as cation. But what is the anion in $(1)$ and $(2)$ ? Impossible to know presently and to separate ($1$) and $(2)$. Which one is $\ce{AgNO3}$ and which one is $\ce{Ag2SO4}$? We'll discover it later on.
Now we will heat all mixtures and try to discover which ones smell ammonia when hot. This will happen in mixtures containing $\ce{NH4+ + OH-}$. This odor is observed in mixtures $(3)$ + $(8)$, $(3)$ + $(9)$, $(6)$ + $(8)$ and $(6)$ + $(9$). As we know that $(9)$ is $\ce{NH4I}$, we know that $(3)$ and $(6)$ are hydroxides. Impossible to separate $\ce{NaOH}$ and $\ce{Ba(OH)2}$ presently. But we see that $(8)$ and $(9)$ contain a cation ammonium $\ce{NH4+}$. As $(9)$ is $\ce{NH4I}$, $(8)$ must be $\ce{NH4Cl}$
Now we will look at mixtures using (7). It is worth noticing that $(7)$ produce only two (white) precipitates with unknown substance : $(3)+(7)$, and $(10)$ + $(7)$. As $(10$) is $\ce{Pb(NO3)2}$ and $(3)$ is either $\ce{NaOH}$ or $\ce{Ba(OH)2}$, we might assume that $(3)$ is $\ce{Ba(OH)2}$ to produce $\ce{BaSO4}$ when mixed with $(7)$. As a conséquence, $(6)$ is $\ce{NaOH}$.
Now, $7$ solutions are known : $(3)$ = $\ce{Ba(OH)2}$, ($5$) = $\ce{KI}$, $(6) = \ce{NaOH}$, $(7) = \ce{Na2SO4}$, $(8) = \ce{NH4Cl}$, $(9) = \ce{NH4I}$. $(10)$ = $\ce{Pb(NO3)2}$
To sépare $(1)$ from $(2)$, we mix both with $(3)$. Only $(2)$ makes a precipitate with $(7)$. So $(2)$ is $\ce{Ag2SO4}$ and $(1)$ is $\ce{AgNO3}$.
The last solution to be analyzed is $(4)$ which, by elimination, must be $\ce{KCl}$. It would make precipitates when mixed with silver salts $(1)$ and $(2)$, and maybe with $(10)$, if the concentration of $\ce{PbCl2}$ is sufficient in the mixture (higher than $0.67$ g in $100$ mL = $0.0024$ M.