This is the bijection indicated in the answer by Matt Fayers, described more explicitly. Note that I needed to correct the wrong definition of "conormal nodes" that I had found in the document mentioned in my comment to that answer.

Represent partitions by their Young diagram, laid out on a checkerboard pattern that makes the top-left square a black one. Condition (3) means that the diagram covers equally many black and white squares, while conditions (1) and (2) mean that all nonzero column lengths are different.

The procedure below will successively add $1$ black square, then $2$ white squares, then $3$ black squares, and so on with alternating colours and increasing numbers, until at some point a step adds squares to every nonempty column. At that point, if there are now $k$ nonempty columns (and the last step added $k$ squares), remove $k,k-1,\ldots,2,1$ squares respectively from the successive columns (which compensates for all the the squares that had been added); this makes the column lengths (no longer necessarily distinct but) all even (this follows from the colour of the last square added to each column), and dividing them all by $2$ gives the desired partition (or its conjugate). The crux of the definition of the procedure is of course to specify exactly which $i$ squares to add in step $i$; this is done as follows.

From the theory of $2$-quotients, it follows that for any Young diagram, if $d$ is the number of black squares it contains minus the number of white squares, then the number of places (cocorners) where a white square can be added minus the number of places (corners) where a white square can be removed equals $2d$, and then the number of places where a black square can be added minus the number of places where a black square can be removed equals $1-2d$. An elementary proof would be to check it for $2$-cores (staircase diagrams), and check that the mentioned differences do not change when adding a domino to the diagram. In the above description, where the number $d$ takes successive values $0,1,-1,2,-2,\ldots$, one sees that at the point where one must add $i$ squares of some colour, the number of addable squares of that colour exceeds the number of removable squares of that colour by $i$ (so in particular one cannot have a shortage of candidate squares). Now pair up each such removable square with an addable square which it "cancels", as follows. Order all these addable and removable squares of the correct colour cyclically, from the top-right to the bottom left along the border of the diagram, and then "warp around" back to the start. If a removable square pair is immediately followed in this order by an addable square then the two pair up; continue pairing up with the same rule, but ignoring already paired-up squares for the "immediately followed" condition. All removable squares are eventually paired up, and $i$ addable squares remain unpaired. It is these $i$ squares that will be effectively added in this step.

(One can avoid talking about a "wrap around" cyclic order, in which case some number $l$ of removable squares may be left unpaired; then one only adds the $i$ *lowest* unpaired addable squares and $l$ addable squares before them are not added, although they remained unpaired. However I find it more intuitive to say that every removable square knocks out precisely one addable square, and without cycling there are precisely $l$ unmatched removable squares, which are all below all unmatched addable squares.)

The proof that each step is reversible is easy. Each square that was added has become removable; therefore the pairing for the reverse direction proceeds by having each addable square pair up with a removable square immediately preceding it in the cyclic order (again ignoring previously paired-up squares), and it is the unpaired removable squares that in the end are effectively removed.

Of course one must also show that the construction in both directions preserves the "distinct nonzero column lengths" condition; this is also easy. In the forward direction, one could only make two column lengths equal if both columns had addable squares of the correct colour, but only the rightmost one of them was actually added. But that means the leftmost one must have been paired up with a removable square; but such pairs are never in the same column, and the presence of the rightmost addable square makes such a pairing impossible. The argument in the oppsite direction is quite similar.