Phenolphthalein is only showing whether if the solution is basic or not and its color becomes darker purple when its ph value is around 14. if pH value is lower than 7, it will not change the color of the solution, so you cannot tell whether if its ph value is close to 0 or 6.
We also know that ph values of 0.1M NaoH and HCI are 13 and 1 respectively. ph value of 1M and 10M NaOH and HCI are 14 and 0 respectively. In other words, you cant tell anything about the mix if the concentration is over 1M.
Lastly, it is given that Phenolphthalein only exists in acidic solutions.
With the information above,
We will take 5 random small samples (let's say 1x amount from each) from 6 solutions and mix them and possible color results will be like the Table 1.
If the color does not change, it means the last solution $f$ which is not mixed is basic for sure, if its color is purple, it means $f$ is acidic. Then use $f$ with the given amount of sample in the table below on the right hand side, mixing it with each solution ($a,b,c,d,e$) by 1x. For example, if the color of the mix is purple, we will take 0.99x of $f$ and 1x of $a$ to test the color. It will take 6 titrations until now.
As a result, the color table will be like below:
From now on, we need to act for each type of result we get one by one if necessary for some cases. For example (first line and worst case condition):
First one result is no color or yellow in the table. That means $f$ is basic as you see. and there will be 3 purple after mixing it with $a,b,c,d$ and $e$. So we will know that $a,b,c$ are acidic, and $d,e$ is basic. Moreover, by using the table, we can easily conclude that $f$ is 10M NaOH! By using this information, we can determine each acidic solution Molars by adding like very small increment from 0.1x, 0.1000001x of 10M NaOH (which is practically like adding 1x of 1.000001M solution into each acid sample) to each acidic new taken samples and the result becomes like on the table below for acidic solutions and we will find out by color of each acidic solution molars (3 more titrations, actually it is 2 but cant know OP would be okay with that by knowing the solution by just color instead of color difference since one of them is such as 0.000001M and the other one is 0.9M, it will be very light purple for one of them, the other one will be dark purple so I would say 3 for now), lastly for two basics, it is very easy by using the acidic samples, just put 0.5x of 1M of HCI into a random basic solution. If its color changes to purple it is 1M NaOH and its color does not change, it is 0.1M NaoH and the other one will be known automatically. So it is 1 more titration)
it requires $10$ titrations/mixing (9 if OP will allow color difference will be noticable by experimenter) to find each solution if 5 sample solution mixing result is colorless. Actually the second part where the color becomes purple after mixing 5 of them is the same procedure and simpler since we know $f$ is acidic and contains phenolphthalein, will take at most 9 titrations at most so I do not want to exemplify for each condition. I can if you ask.