One more a bit “particular” answer, however I’d say it isn’t in opposition to the (present) guidelines, it is only a matter of interpretation.

1.

$ VII = 111 $
Roman $7$ = Binary $7$

2. and one other one…

$$ | -11 | = 11 $$
$$ abs(-11) = 11 $$

3.
…and one other one…

$ II = 10 $
Roman $2$ = Binary $2$

4.
… and one other “new” idea…

$$-iiii = -1$$
with the imaginary unit $i^2 = -1$
$$-i*i*i*i = -1$$

5.
…though the “recreation is over”…

…a brand new mixture from recognized rules…

Will be interpreted as
$|1| = |-1|$ or $|i| = |-i|$ or $|1| = |-i|$ or $|i| = |-1|$

6.
… and a barely “odd” one … (possibly I ought to cease now…)

$ XI = +11 $
Roman $11$ = $+11$

7.
… one more unconventional “rot90” model…

Roman $2$ = 90° rotated $2$

8.
… one other one, possibly a bit too nerdy or … ?

$ 11 = 11 | 11 $
in lots of programming languages $|$ is an emblem for bitwise inclusive OR
so, this may be interpreted as
binary $3$ = binary $3$ OR binary $3$
or
decimal $11$ = decimal $11$ OR decimal $11$

9.
… one other mixture of rules from above…

$ 11 = omega $
Binary $3$ = 90° clockwise rotated 3

10.
… sorry, once more… a easy one which isn’t but listed in any of the solutions…

$ II – 1 = +1 $
Roman $2$ – $1$ = $+1$

11. … and a final(?) one…

$ II + 1 = 11$
Roman $2$ + $1$ = Binary $3$

12.
… another bases for brand spanking new choices…

(base 4) $11 = 5 $ (rotated by 90°)
$11_4 = 5 $

13.
… and mixed with earlier ones…

$ VIII = 11 $ (base 7)
Roman $8 = 11_7 $

hmmm, I could not resist…

14. …on your information(rmatics)…

$ 7 wedge 11 = 1 $
$wedge$ is used as binary unique OR (XOR)
$7_{10} wedge 11_5 = 1$
$ 7_{10} wedge 6_{10} = 1 $
$ 111_2 wedge 110_2 = 1 $

15. … a particular transfer…

$ 71 – 1 = 11 $
$ 71_{10} – 1 = 11_{69} $ (base 69, okay very particular)

16.
…seven & eleven fo(u)r your convencience …

$ 11 – 7 = 11 $
$ 11_{10} – 7_{10} = 11_3 $
$ 4 = 4_{10} $

17. … and eventually(?) a nicer one…

(binary) $ 1111 = F$ (hexadecimal)
$1111_2 = F_{16}$
$15 = 15$

18.- 30.
…a Roman’s ocean of elevens…

Really, $11_n = (n+1)_{10}$. Therefore $11$ may be interpreted as any decimal quantity $>=3$ relying on the bottom $n$. So, this boils all the way down to which numbers may be created on the opposite facet having 5 matches with precisely 2 strikes.
You are able to do this for instance with the next Roman numbers:
$$ VIII, XIII, XIV, XVI, XIX, XXI, LIV, LVI, LIX, LXI, CII, DII, M $$
Really, $VIII$ has been used already above, however you could possibly additionally use some much less frequent writings of roman numbers, e.g. $IIC$.
For illustration, simply the final 3 variations.

31. – 33.
… 111-Hattrick… some good $111$-solutions ought to be talked about:

$$111_3 = 1D_{16} rightarrow 13_{10} = 13_{10} $$
$$111_5 = 1F_{16} rightarrow 31_{10} = 31_{10}$$
$$111_{15} = F1_{16} rightarrow 241_{10} = 241_{10}$$

34. – $infty$

For the association $11111_m = 11_n$ you may mainly discover an infinite variety of appropriate bases for $m$ and $n$.

Now, I suppose it’s sufficient… except any individual needs extra.