Normal and visual string reversion

05AB1E, 16 bytes

Uses the fact that 05AB1E has a constant predefined to "()<>[]{}" and isn't affected the visually reversion.

Code:

,q‡"}{][><)("užR

Explanation:

,                 # Pop and print the input.
 q                # Quit.
  ‡"}{][><)("užR  # This part is ignored.

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Reversed:

Ržu"()<>[]{}"‡q,

Explanation:

R                 # Reverse the input.
 žu               # Short for "()<>[]{}".
   "()<>[]{}"     # Push this string.
             ‡    # Transliterate (no-op, since everything is transliterated to itself).
              q   # Quit and implicitly print.
               ,  # This part is ignored.

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Visually reversed:

Ržu")(><][}{"‡q,

Explanation:

R                 # Reverse the input.
 žu               # Short for "()<>[]{}".
   ")(><][}{"     # Push this string.   
             ‡    # Transliterate (giving the visually reversed string).
              q   # Quit and implicitly print.
               ,  # This part is ignored.       

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Uses CP-1252 encoding.

CJam, 21 bytes

qe#ere$_"}{][><)("%Wq

Test it here.

Normal reversion:

qW%"()<>[]{}"_$ere#eq

Test it here.

Visual reversion:

qW%")(><][}{"_$ere#eq

Test it here.

Explanation

First, the normal code:

qe#ere$_"}{][><)("%Wq

This is simple: q reads all input, e# comments out the remainder of the program, and the input is printed implicitly at the end.

Now the normal reversion:

q            e# Read all input.
W%           e# Reverse it.
"()<>[]{}"   e# Push this string.
_$           e# Duplicate and sort it. However, the string is already sorted
             e# so we just get two copies of it.
er           e# Transliteration (i.e. character-wise substitution). But since the
             e# source and target string are identical, the reversed input
             e# is left unchanged.
e#eq            Just a comment...

And finally, the visual reversion:

q            e# Read all input.
W%           e# Reverse it.
")(><][}{"   e# Push this string.
_$           e# Duplicate and sort it. This gives us "()<>[]{}", i.e. the
             e# same string with each bracket pair swapped.
er           e# Transliteration (i.e. character-wise substitution). This
             e# time, this toggles all the brackets in the reversed input
             e# completing the visual reversion.
e#eq            Just a comment...

Haskell, 124 bytes

Forward:

f=id
--esrever.q pam=2>1|esrever=2<1|f;x=x q;')'='(' q;'('=')' q;']'='[' q;'['=']' q;'>'='<' q;'<'='>' q;'}'='{' q;'{'='}' q

Normal reverse:

q '}'='{';q '{'='}';q '>'='<';q '<'='>';q ']'='[';q '['=']';q ')'='(';q '('=')';q x=x;f|1<2=reverse|1>2=map q.reverse--
di=f

Visual reverse:

q '{'='}';q '}'='{';q '<'='>';q '>'='<';q '['=']';q ']'='[';q '('=')';q ')'='(';q x=x;f|1>2=reverse|1<2=map q.reverse--
di=f

Each version defines a function f which takes and returns a string. In forward mode f is the identity function id, the rest of the code is a comment. In normal reverse mode the guard 1<2 in f is True, so reverse is applied. In visual reverse mode, the < is switched to > and the guard is False. The second guard is just the other way around and Truein visual mode, so additionally q is applied which switches "()<>{}[]".

f|1<2=reverse|1>2=map q.reverse      -- normal reverse mode
f|1>2=reverse|1<2=map q.reverse      -- visual reverse mode

Besides <and > in the guards, my code doesn't use any of the brackets, so they can't be messed up.