Identify each substances

Put a little amount of each substance into each testing tube and number each tube (to keep track of what is in each tube)

Then, pour excessive $NaOH$ solution into each tube and let all reactions happen. Observe all reactions. There will be some precipitates with different colors.

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The tube containing a red-brown precipitate must have contained $FeCl_3$ previously.

$FeCl_3 + 3NaOH \rightarrow Fe(OH)_3 \downarrow + 3NaCl$

Note that $Fe(OH)_3$ is red-brown.

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The tube containing a blue precipitate must have contained $CuCl_2$ previously.

$CuCl_2 + 2NaOH \rightarrow Cu(OH)_2 \downarrow + 2NaCl$

Note that $Cu(OH)_2$ is blue.

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The tube containing a white precipitate must have contained $MgCl_2$ previously.

$MgCl_2 + 2NaOH \rightarrow Mg(OH)_2 \downarrow + 2NaCl$

Note that $Mg(OH)_2$ is white.

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The tube containing a white precipitate, which is then dissolved in leftover $NaOH$ solution must have contained $AlCl_3$ previously.

$AlCl_3 + 3NaOH \rightarrow Al(OH)_3 \downarrow + 3NaCl$

Note that $Al(OH)_3$ is white. Then, it's dissolved in leftover $NaOH$ solution: $Al(OH)_3 + NaOH \rightarrow NaAlO_2 + 2H_2O$

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The remaining tube contains $NaCl$ .

Therefore, we can use $\boxed{NaOH}$ only to identify all substances.

Bonus: Any strong base solutions from alkaline metals (Na, K, Ca, Ba,...) can also be used.