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14.

15.

16.

$ 11 - 7 = 11 $
$ 11_{10} - 7_{10} = 11_3 $

17.

 ... and finally(?) a nicer one...

Actually, $11_n = (n+1)_{10}$. Hence $11$ can be interpreted as any decimal number $>=3$ depending on the base $n$. So, this boils down to which numbers can be created on the other side having 5 matches with exactly 2 moves.
You can do this for example with the following Roman numbers:
$$ VIII, XIII, XIV, XVI, XIX, XXI, LIV, LVI, LIX, LXI, CII, DII, M $$

31. - 33.  ... 111-Hattrick... some nice $111$-solutions should be mentioned:

14. ...for your info(rmatics)...

15. ... a special move...

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

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

17.  ... and finally(?) a nicer one...

Actually, $11_n = (n+1)_{10}$. Hence $11$ can be interpreted as any decimal number $>=3$ depending on the base $n$. So, this boils down to which numbers can be created on the other side having 5 matches with exactly 2 moves.
You can do this for example with the following Roman numbers:
$$ VIII, XIII, XIV, XVI, XIX, XXI, LIV, LVI, LIX, LXI, CII, DII, M $$ Actually, $VIII$ has been used already above, but you could also use some less common writings of roman numbers, e.g. $IIC$.
For illustration, just the last 3 versions.

31. - 33. 
... 111-Hattrick... some nice $111$-solutions should be mentioned:

2.

And another one...

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

2. and another one...

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

18.- 30. ...a Roman's ocean of elevens...

Actually, $11_n = (n+1)_{10}$. Hence $11$ can be interpreted as any decimal number $>=3$ depending on the base $n$. So, this boils down to which numbers can be created on the other side having 5 matches with exactly 2 moves.
You can do this for example with the following Roman numbers:
$$ VIII, XIII, XIV, XVI, XIX, XXI, LIV, LVI, LIX, LXI, CII, DII, M $$

31. - 33. ... 111-Hattrick... some nice $111$-solutions should be mentioned:

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

34. - $\infty$

For the arrangement $11111_m = 11_n$ you can basically find an infinite number of suitable bases for $m$ and $n$.

$ 7 \wedge 11 = 1 $
$\wedge$ is used as binary exclusive OR (XOR)
$7_{10} \wedge 11_5 = 1$

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