Scenario: This looks specifically at a near-miss scenario with an Earth-like intruder passing by at a distance of 1,000,000 km at a speed of 100 km/s. This is about three times the distance to the Moon and at about three times Earth’s orbital velocity. The 100 km/s seems a reasonable number given the relative velocities of our stellar neighbors, the solar infall/escape velocity, and the likely difference in directions of the Earth and intruder velocities.
First Sighting: It's a pure guess, but maybe the intruder will be first recognized when it’s brightness (apparent magnitude) becomes similar to that of Pluto. The difference in absolute magnitude between the Earth and Pluto is (-3.99)-(-0.70) = -3.29. This corresponds to a difference in brightness of 100^(3.29/5) = 20.7. At that distance from the Earth/Sun, the brightness will go as the 4th power of distance, so we should first spot the intruder when it is 20.7^(1/4) = 2.13 times further away than Pluto or about 12.6 billion km. Just approximating with a constant 100 km/s, this means we will get about 4 years of advance warning.
Night Sky: Assuming the intruder approaches from roughly the opposite direction of the Sun, it will be visible in the night sky with a magnitude -3.99+(1/5)log100(1/s^2/(1+s)^2 where s is the distance from Earth in AU. At 0.6 AU (90,000,000 km) it will be roughly as bright as Venus and clearly visible in the night sky. This will occur about 10 days before closest approach.
At its closest approach (1 million km), the intruder will appear about 40% larger diameter than the Moon and will be about 7 times brighter, both because of the larger apparent area (0.73/0.52)^2 and higher albedo (0.43/0.12). The intruder will take 5.6 hours to travel the 2 million kilometers across 90 degrees of the night sky while its apparent area goes from 50% to 100% and back to 50% of maximum. A sight worth watching!
Orbital Disturbance (Earth): Assuming the deflections of the Earth and the intruder are small, the delta_V can be calculated as (1/2)GMe/d^2 x (2d/v), where Me is the mass of the Earth or the intruder, d is the distance of closest approach and v is the velocity. The first factor corresponds to the maximum acceleration of the Earth and the second factor to the effective time during which the acceleration is applied. The first factor is 0.000199 m/s^2 and the second factor is 20,000 seconds giving a delta_V of 3.98 m/s. This is 0.00013 of the Earth’s orbital speed of 30 km/s. Generally this would change the ellipticity of the orbit slightly and could also change the mean radius if the delta-V were mainly along the direction of the orbit. The orbital energy could change by +/-0.00026 and the mean radius would change the same fraction.
Corresponding to a change in the Earth’s mean orbital radius, the insolation would change by double that fraction or +/-0.00052. The temperature based on black body radiation would change by the fourth root of the insolation, i.e. +/-0.00013 or +/-0.013% or +/-0.04 degrees C. These changes seem quite minor.
Orbital Disturbance (Moon): The Moon would be more severely affected because its mass is small compared to the intruder, so the acceleration would be roughly twice as large. Also the Moon could be on the ‘wrong’ side where it would be closer to the intruder by a factor of 2/3. So the acceleration and the delta_V could be higher by a factor of 2/(2/3)^2 = 4.5. The resulting delta_V of 18 m/s on the Moon’s orbital velocity of about 1000 m/s is more significant. This could result in a 3.6% change in orbital energy and in radius and a change of 5.4% in period, changing the 28 day period by more than a day. Since tidal effects go as the cube of distance, presumably the lunar tidal ranges would increase (or decrease) by about 11%.
Transient tidal effect: Tidal forces are proportional to mass/distance^3. Ignoring the constant factor, the peak tidal force from the Moon : Sun : Intruder are in the ratio 7.35E22/(3.84E5)^3 : 1.989E30/(1.48E8)^3 : 5.97E24/(1E6)^3 or 1.30 : 0.61 : 5.97. So the maximum tidal force as the intruder swings by will be about three times more that the combined lunar and solar tidal force. Presumably, during the several hours of closest approach, this will raise ocean tides about three times larger than the familiar large spring tides. Coastal areas would experiences tides tens of feet higher than normal much as in a storm surge or tsunami. I suppose it is also possible that some latent earthquakes could also be released by such large tidal forces
(The last time I did a back-of-the-envelope calculation like this, I was wrong by a factor of 10^6. But it certainly elicited some pretty quick responses!)