This is a follow-up question to: Does a symmetry necessarily leave the action invariant?

Qmechanic writes here:

Here the word off-shell means that the Lagrangian eqs. of motion are not assumed to hold under the specific variation. If we assume the Lagrangian eqs. of motion to hold, any variation of the Lagrangian is trivially a total derivative.

Qmechanic writes here:

if an action (1) has a quasi-symmetry, then the EOM (2) must have a symmetry (wrt. the same transformation).

1. What exactly is an off-shell symmetry? I'm now confused. Does it mean that the action changes by a boundary term but despite that, the transformation does not necessarily map a solution of the EOM to a solution of the EOM? That seems to contradict the second quote---or does it?

2. What is the proof of the "trivial" fact that for an on-shell symmetry, the Lagrangian necessarily changes by a total derivative?

This is a follow-up question to: Does a symmetry necessarily leave the action invariant?

Qmechanic writes here:

Here the word off-shell means that the Lagrangian eqs. of motion are not assumed to hold under the specific variation. If we assume the Lagrangian eqs. of motion to hold, any variation of the Lagrangian is trivially a total derivative.

Qmechanic writes here:

if an action (1) has a quasi-symmetry, then the EOM (2) must have a symmetry (wrt. the same transformation).

1. What exactly is an off-shell symmetry? I'm now confused. Does it mean that the action changes by a boundary term but despite that, the transformation does not necessarily map a solution of the EOM to a solution of the EOM? That seems to contradict the second quote---or does it?

2. What is the proof of the "trivial" fact that for an on-shell symmetry, the Lagrangian necessarily changes by a total derivative?

This is a follow-up question to: Does a symmetry necessarily leave the action invariant?

Qmechanic writes here and here:

Here the word off-shell means that the Lagrangian eqs. of motion are not assumed to hold under the specific variation. If we assume the Lagrangian eqs. of motion to hold, any variation of the Lagrangian is trivially a total derivative.

But also

if an action (1) has a quasi-symmetry, then the EOM (2) must have a symmetry (wrt. the same transformation).

1. What exactly is an off-shell symmetry? I'm now confused. Does it mean that the action changes by a boundary term but despite that, the transformation does not necessarily map a solution of the EOM to a solution of the EOM? That seems to contradict the second quote---or does it?
2. What is the proof of the "trivial" fact that for an on-shell symmetry, the Lagrangian necessarily changes by a total derivative?

This is a follow-up question to: Does a symmetry necessarily leave the action invariant?

Qmechanic writes here:

Here the word off-shell means that the Lagrangian eqs. of motion are not assumed to hold under the specific variation. If we assume the Lagrangian eqs. of motion to hold, any variation of the Lagrangian is trivially a total derivative.

Qmechanic writes here:

if an action (1) has a quasi-symmetry, then the EOM (2) must have a symmetry (wrt. the same transformation).

1. What exactly is an off-shell symmetry? I'm now confused. Does it mean that the action changes by a boundary term but despite that, the transformation does not necessarily map a solution of the EOM to a solution of the EOM? That seems to contradict the second quote---or does it?

2. What is the proof of the "trivial" fact that for an on-shell symmetry, the Lagrangian necessarily changes by a total derivative?