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Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit.

You've neglected to account for the magnetic attraction as the pendulum bob goes back to its central position.

On the outwards leg, you are correct that the magnet's attraction will pull on the bob and give it more energy than it would have in the absence of the magnet. However, in the return leg, the pendulum bob is trying to get away from the magnet's attractive force, and this will claim back all of the additional energy.

(... if the system is perfect, that is. Real-world magnetic materials will show some amount of hysteresis, so the bob will lose slightly more energy on the way back than it gained on the way out.)

This type of mistake is quite common when you have a core dynamics which is known to be conservative, and still seems to be producing energy - you're just conveniently neglecting to take into account the parts of the cycle where that force performs work against your system. For a similar example in action, see What prevents this magnetic perpetuum mobile from working?.

Perpetual motion is impossible due to dissipation, or if you prefer second principle of thermodynamics and not to energy conservation.

If your analysis of the suggested setup was correct, you could create mechanical energy for free !

In first analysis, neglecting the (unavoidable) losses in the ferromagnetic medium, your system is conservative : what you have is a modified pendulum, where the confinement potential contains not only the gravitational part but also a magnetic component. Actually the magnetic force slightly decrease the recall torque that you would have with gravity alone, and the amplitude of motion will indeed be larger. But you nevertheless will have a turning point where the kinetic energy vanishes, and when going back, you will reach exactly the same angle for the turning point on the other side.
This correspond to an oscillator with constant amplitude, because losses have been neglected.
The sources of losses are at least : friction in air, friction on axe, ferromagnetic hysteresis, Foucault currents. So the amplitude will decrease and the perpetual motion be reduced to perpetual immobility...

You are missing that as the metal passes through the changing magnetic field, Eddy currents will be produced.

These currents will result in the metal heating up (with the amount depending on the speed of travel and strength of field).

This heating essentially results in energy being removed from the system; and thus the bobbing can not be done forever; even if the rest of the system is perfect.

A magnet strong enough to pull the metal up will be too strong to let it fall again.

If you would use static magnets for that, than while swinging back, the metal is closer to the magnet attracting it in opposite direction relative to its movement and thus takes more momentum away as it gains from the other magnet. Thus the next amplitude is lower. The result is, that your amplitude overall would even be damped relative to the free pendulum.

The accelerating force by the magnet on the far side is proportional to $cfrac{1}{(d+sin phi l)^2}$ and the deccelerating force by the magnet on the close side is proportional to $cfrac{1}{(d-sin phi l)^2}$.

Where d is the distance between pendulum mass and either magnet in rest state, $phi$ is the deflection angle and l is the pendulum length. Since the Magnets are identical, they carry the same prefactor. Thus resulting force adding to the positively accelerating gravitational force is proportional to

$$ cfrac{1}{(d+sin phi l)^2} - cfrac{1}{(d-sin phi l)^2} = cfrac{(d-sin phi l)^2 - (d+sin phi l)^2}{(d-sin phi l)^2 (d+sin phi l)^2} $$

As the denominator is now again the same for both contributing forces we can neglect it for the ongoing computation and result in the total magnetic force proportional to:

$$ (d-sin phi l)^2 - (d+sin phi l)^2 = d^2 - 2dsin phi l + sin^2 phi l^2 - d^2 - 2dsin phi l - sin^2 phi l^2 = -4dsin phi l $$

thus the oscillation given by the gravitational force is damped by the magnetic forces.

Edit: My original train of thoughts was wrong on this one. At least the calculations show that the oscillator is driven by a smaller force compared to the free oscillator (just gravity). Thus without any energy loss the 'magnetic oscillator' would just oscillate slower without any change in amplitude.