Splitting a shape into a number of congruent shapes similar to the original; analogues to A4 paper system Two A4 papers can be put side by side to create a new rectangle that is similar to the original rectangles. 
Similarly, 5 right triangles with the length ratio between legs of 1:2 can be put to create a new triangle with the same 1:2 ratio as follows:

Is the same can be said for 3? i.e. is there a ‘shape’ which can be split into 3 congruent shapes that are similar to the original?
It seems to me that some parameterization of the L-shape could possibly work.

More generally, for any $n$, does there always exist a shape which can be tiled by $n$ congruent pieces similar to itself?
Remark : I don’t know how to define what constitutes a ‘shape’ so if you want to be specific it could be a ‘connected’ polygon.
EDIT: Ronald Blaak has already provided a simple answer to this though I would greatly appreciate if anyone could give some unusual examples of shapes that can tile itself. 
 A: Yes, the simplest solution is a rectangle.
Consider a rectangle with sides $a < b$ and place $n$ of them alongside to obtain a rectangle with sides $b$ and $na$. Then the requirement would be
$$
a : b = b : n a
$$
which is satisfied by $b = \sqrt{n} a$.
Simple solutions for square numbers $n$ can be obtained by taking an arbitrary triangle or parallelogram and drawing regular distributed lines parallel to the sides and find a square number of smaller copies inside. In fact you make use of that in the first example, the four lower left triangles do just that.
Also the L-shape figure you show, if it is formed by three joined squares, can be used. Also in this case a square number of these shapes can always be fitted in a larger similar shape.
In a similar fashion an L-shape of length 3 and width 2 (formed by 4 unit-squares) can be used. 
The $n=5$ case is a rather special one and requires particular ratio of the sides of a right angled triangle. The same trick can however be used to obtain other values of $n$ that are not squares if we use right angled triangles as the basic unit. I.e. take right angled triangle with sides $a<b<c$. Using the regular grid mentioned above one can create larger versions of $k^2$ and $l^2$ units ($k<l$). These congruent triangles can be placed together to form an congruent larger triangle provided $k b = l a$, so when $b = (l/k)a$ and results in a triangle consisting of $n=k^2+l^2$ smaller ones. 
