Charcoal, 50 43 bytes

Ｆ³²⁴«Ｐ++↷ＡＥ…¹¦⁵∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Try it online! Link is to verbose version of code. I originally tried various reflections and rotations but they either didn't do what I want or in some cases were buggy. I then tried a nested loop approach but I've now switched to this iterative method which works by drawing a number of lines between each inner cross depending on how many powers of 3 the step number is divisible by. It can even be readily modified to accept a size parameter at a cost of only 4 bytes:

ＮβＦ×⁴Ｘ³β«Ｐ++↷ＡＥ…·¹β∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Edit: I've since worked out how to use RotateShutterOverlap to achieve this task, but annoyingly it takes me 44 bytes:

Ａ⁰ηＦ⁶«ＡηγＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ-γ+¿γ⟲ＳＯ²⁶⁻×²γ¹»‖⟲ＳＯ⁹⁵

If RotateShutterOverlap accepted a variable rotations parameter, that would reduce it to 40 bytes:

Ａ⁰ηＦ⁶«Ａ∨η¹γＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ+γ+⟲ＳＯ⎇‹ι⁵Ｌβ²⁴⁶γ

Charcoal, 50 43 bytes

Ｆ³²⁴«Ｐ++↷ＡＥ…¹¦⁵∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Try it online! Link is to verbose version of code. I originally tried various reflections and rotations but they either didn't do what I want or in some cases were buggy. I then tried a nested loop approach but I've now switched to this iterative method which works by drawing a number of lines between each inner cross depending on how many powers of 3 the step number is divisible by. It can even be readily modified to accept a size parameter at a cost of only 4 bytes:

ＮβＦ×⁴Ｘ³β«Ｐ++↷ＡＥ…·¹β∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Edit: I've since worked out how to use RotateShutterOverlap to achieve this task, but annoyingly it takes me 44 bytes:

Ａ⁰ηＦ⁶«ＡηγＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ-γ+¿γ⟲ＳＯ²⁶⁻×²γ¹»‖⟲ＳＯ⁹⁵

If RotateShutterOverlap accepted a variable rotations integer, that would reduce it to 40 bytes:

Ａ⁰ηＦ⁶«Ａ∨η¹γＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ+γ+⟲ＳＯ⎇‹ι⁵Ｌβ²⁴⁶γ

As it is, using a rotations list parameter takes 45 bytes:

Ａ⁰ηＦ⁶«Ａ∨η¹γＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ+γ+⟲ＳＯ⟦⁶ײ⁺¹⁼⁵ι⟧⁻ײγ¹

Charcoal, 50 43 bytes

Ｆ³²⁴«Ｐ++↷ＡＥ…¹¦⁵∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Try it online! Link is to verbose version of code. I originally tried various reflections and rotations but they either didn't do what I want or in some cases were buggy. I then tried a nested loop approach but I've now switched to this iterative method which works by drawing a number of lines between each inner cross depending on how many powers of 3 the step number is divisible by. It can even be readily modified to accept a size parameter at a cost of only 4 bytes:

ＮβＦ×⁴Ｘ³β«Ｐ++↷ＡＥ…·¹β∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Edit: I've since worked out how to use RotateShutterOverlap to achieve this task, but annoyingly it takes me 44 bytes:

Ａ⁰ηＦ⁶«ＡηγＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ-γ+¿γ⟲ＳＯ²⁶⁻×²γ¹»‖⟲ＳＯ⁹⁵

Charcoal, 50 43 bytes

Ｆ³²⁴«Ｐ++↷ＡＥ…¹¦⁵∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Try it online! Link is to verbose version of code. I originally tried various reflections and rotations but they either didn't do what I want or in some cases were buggy. I then tried a nested loop approach but I've now switched to this iterative method which works by drawing a number of lines between each inner cross depending on how many powers of 3 the step number is divisible by. It can even be readily modified to accept a size parameter at a cost of only 4 bytes:

ＮβＦ×⁴Ｘ³β«Ｐ++↷ＡＥ…·¹β∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Edit: I've since worked out how to use RotateShutterOverlap to achieve this task, but annoyingly it takes me 44 bytes:

Ａ⁰ηＦ⁶«ＡηγＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ-γ+¿γ⟲ＳＯ²⁶⁻×²γ¹»‖⟲ＳＯ⁹⁵

If RotateShutterOverlap accepted a variable rotations parameter, that would reduce it to 40 bytes:

Ａ⁰ηＦ⁶«Ａ∨η¹γＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ+γ+⟲ＳＯ⎇‹ι⁵Ｌβ²⁴⁶γ

Charcoal, 50 43 bytes

Ｆ³²⁴«Ｐ++↷ＡＥ…¹¦⁵∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Try it online! Link is to verbose version of code. I originally tried various reflections and rotations but they either didn't do what I want or in some cases were buggy. I then tried a nested loop approach but I've now switched to this iterative method which works by drawing a number of lines between each inner cross depending on how many powers of 3 the step number is divisible by. It can even be readily modified to accept a size parameter at a cost of only 4 bytes:

ＮβＦ×⁴Ｘ³β«Ｐ++↷ＡＥ…·¹β∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Edit: I've since worked out how to use RotateShutterOverlap to achieve this task. This version takes 50 bytes and can be parameterised by replacing the 5 with InputNumber():

Ａ¹γ+Ｆ…·γ⁵«⟲ＳＯ²⁶γＡ÷×Ｘ²ιι⁸γＪ⁻Ｘ²ιι⁰γ+γＡ⁺ײγ¹γ»⟲ＳＯ²⁴⁶γ

This version takes 49 bytes but should really be shorter; I think I must keep running into bugs in Charcoal, because my variables seem to change unexpectedly:

Ａ⁰ηＦ⁶«ＡηγＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ-γ+¿γ⟲ＳＯ²⁶⁻×²γ¹»Ｊ⁹⁵¦⁰+⟲ＳＯ⁹⁵

(For those interested, Ａ⁰ηＦ⁶«ＡηγＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ-γ+¿γ«Ａ⁻ײγ¹γ⟲ＳＯ²⁶γ»»Ｊγ⁰+⟲ＳＯγ throws an unexpected exception.)

Charcoal, 50 43 bytes

Ｆ³²⁴«Ｐ++↷ＡＥ…¹¦⁵∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Try it online! Link is to verbose version of code. I originally tried various reflections and rotations but they either didn't do what I want or in some cases were buggy. I then tried a nested loop approach but I've now switched to this iterative method which works by drawing a number of lines between each inner cross depending on how many powers of 3 the step number is divisible by. It can even be readily modified to accept a size parameter at a cost of only 4 bytes:

ＮβＦ×⁴Ｘ³β«Ｐ++↷ＡＥ…·¹β∧¬﹪ιＸ³κ⁻Ｘ²⁺κ¹⁺κ²εＦ⁺ε⮌ε¿κ«+κ↶

Edit: I've since worked out how to use RotateShutterOverlap to achieve this task, but annoyingly it takes me 44 bytes:

Ａ⁰ηＦ⁶«ＡηγＡ⁻⁺Ｘ²ιηιηＪη⁰Ｐ-γ+¿γ⟲ＳＯ²⁶⁻×²γ¹»‖⟲ＳＯ⁹⁵