Final solution
Freddy is
51
and the four mathematicians' ages are
51, 66, 70, 85 ($51=3\times17$, $66=2\times3\times11$, $70=2\times5\times7$, $85=5\times17$),
with their Collatz-transformed ages being respectively
154, 33, 35, 256 ($154=2\times7\times11$, $33=3\times11$, $35=5\times7$, $256=2^8$).
Step-by-step deduction
This is essentially a tricky exercise in modular arithmetic and prime factorisations.
There are at most two even numbers in each of the two graphs (three even numbers would give a triangle). But the Collatz transform sends every odd number to an even number, so there are exactly two odd and two even numbers in each graph.
The number in the centre of the post-Collatz graph must be even, since the two even numbers are obviously linked. Every even post-Collatz number must be congruent to 1 mod 3, therefore congruent to 4 mod 6 by the Chinese remainder theorem.
So that central number in the post-Collatz graph must be
congruent to 4 mod 6, having at least three distinct prime factors (to be connected separately to the three others), and at most 300.
Let's consider all possible numbers satisfying these criteria:
$70,130,154,190,220,238,280,286$ (yes, I did find all these by hand!)
Most of these don't work because
either they give a prime number in the pre-Collatz graph ($70=3(23)+1$, $130=3(43)+1$, $220=3(73)+1$, $238=3(79)+1$) or they give a prime factor which is too large ($280=3(3\times31)+1$, something else in the pre-Collatz graph must be a multiple of $31$ which is impossible).
The smallest one which might work is
$154=3(51)+1$. This is the Collatz transform of $51=3\times17$, so we need a multiple of $17$ in the original square. Fortunately, $5\times17=85$ has Collatz transform $3(85)+1=256$, which is a power of 2.
Then for the post-Collatz graph we have
$154=2\times7\times11$ in the middle, $256$, a multiple of 7, and a multiple of 11,
corresponding in the pre-Collatz graph to
$51=3\times17$, $85=5\times17$, a multiple of 3, and a multiple of 5.
Putting the
3 with 11 and 5 with 7,
we get the final answer.
How about uniqueness?
We had three remaining possibilities that might work as the central number in the post-Collatz graph. Now we know
154 does work; what about 190 and 286?
$286=2\times11\times13$ is the Collatz transform of $95=5\times19$. The other multiple of 19 in the pre-Collatz graph must be $3\times19=57$, which Collatz transforms to $172=2^2\times43$. Then the post-Collatz graph is $286$, $172$, a multiple of 11, and a multiple of 13. But one of those last two must also be a multiple of 5, and both $5\times11=55$ and $5\times13=65$ are too big to be Collatz transforms of even numbers less than 100. Contradiction.
$190=2\times5\times19$ is the Collatz transform of $63=3^2\times7$. The post-Collatz graph then must contain the number $19$, corresponding to $38=2\times19$ in the pre-Collatz graph. One of the remaining numbers in the pre-Collatz graph must be $3\times19=57$, which Collatz transforms to $172=2^2\times43$. Then the final number must be $2\times5\times7=70$ pre-Collatz and $5\times7=35$ post-Collatz.
So there is a second possibility, namely Freddy is
38
and the four mathematicians' ages are
38, 57, 63, 70 ($38=2\times19$, $57=3\times19$, $63=3^2\times7$, $70=2\times5\times7$)
with their Collatz-transformed ages being respectively
19, 172, 190, 35 ($19=19$, $172=2^2\times43$, $190=2\times5\times19$, $35=5\times7$).
