We shall consider here the use of squares to draw a perpendicular through a given point to a given line. Full stop.

Obviously, we can then draw parallels (perpendicular to perpendicular), if we accept the use of "auxiliary points",

arbitrary choosen points such that the final construction doesn't depend on the choice of these points.

Choosing an arbitrary point (cyan), a perpendicular to these perpendiculars (hence a parallel to AB) makes a rectangle.

The diagonals intersect at the center of this rectangle, and the perpendicular through I to AB

is the perpendicular bisector of AB, hence intersects AB in its midpoint.

Construct on any perpendiculaire (through arbitrary cyan point) any segment BC and its midpoint M.

A pencil of lines from O, intersect point of AC and MH, allows to complete the construction, for HA/HA' = MC/MB.

for instance we can't copy a segment onto a not parallel line.

Hence we can't construct a square from a given side.

To construct the center of the direct similitude changing AB → A'B'.

Recall the classical construction :

Let I the intersect point of AB and A'B'

The circumcircles to IAA' and IBB' intersect in another point which is the center of the similitude.

(proof from angle property of similitude).

We really can construct this point with straightedge and squares only.

Construct the perpendicular bisectors of IA et IA', intersecting in O,

and also the perpendicular bisectors of IB and IB', intersecting in O'.

We can't draw the circumcircles (centered in O and O'), but the other intersection point is the symmetric S of I with respect to center line OO'.

We already have a segment IA and its midpoint M, just complete the construction of S :

The perpendicular to OO' from I intersects OO' in H.

A parallel to MH from A intersects IH in S.

Well, this construction is unnecessarily complicated.

See below.

The midpoint of a segment can be constructed as above (diagonals of a parallelogram intersects at their midpoints).

But the perpendicular bisector is a little tricky :

Just flip the squares (exchange the role of its sides) to draw angle -θ, hence an isosceles triangle ABO and that's it.

The problem is when θ ≈ 90° for the intersection point of lines at +θ from A and -θ from B is ... very far (at infinity if θ = 90°)

Any parallel to AB makes then an isosceles trapezoid, whose diagonals intersect on the perpendicular bisector.

Drawing two such trapezoids gives two points on the perpendicular bisector.

We can draw perpendiculars even with a false squares !

Points A,B,A',B' can be freely moved.

Let I the intersect point of AB and A'B'

Draw lines at angle θ with AB from A and B, and similarily
the lines at angle θ with A'B' from A' and B'.

Corresponding lines intersect in a, b.
Draw line ab.

Draw a line Is such that angle of ab with Is is θ.

In other words, draw a line from I at angle -θ with ab :
(ab, Is) = - (Is, ab)

Is and ab intersect at the searched similitude center S.

(proof with oriented angles of lines and suggested circles in yellow).

With θ = 90°, this gives a much simpler construction than the first one !

It is then quite easy to draw the transformed from any point M by the similitude :

Draw Mq // AB, intersecting IS in Q.

Draw line Mp at angle θ with Mq (parallel to Aa), intersecting ab in P.

Draw line Qm' // A'B'

Construct the line Pm' at angle θ with Qm' (parallel to A'a)

M' is the intersection point of Qm' with Pm'.

Here also, equal angles and concyclic points M,M',P,Q,S is the key.