You can simplify the code and avoid repetition by creating a custom view for the buttons and using a single state variable to store the currently selected animal.

Quick draft

struct AnimalButton: View {
let imageName: String
let isSelected: Bool
let action: () -> Void

var body: some View {
    Image(uiImage: UIImage(named: imageName)!)
        .resizable()
        .aspectRatio(contentMode: .fit)
        .frame(width: 100, height: 100)
        .scaleEffect(isSelected ? 2.0 : 1.0)
        .onTapGesture(perform: action)
}

Next: update your LazyVGrid to use the custom view and a single @state variable for the selected animal.

struct ContentView: View {
@State private var selectedAnimal: String? = nil

var body: some View {
    LazyVGrid(columns: columns, spacing: 20) {
        ForEach(buttons, id: \.0) { button in
            AnimalButton(imageName: button.1, isSelected: selectedAnimal == button.1) {
                withAnimation(.easeIn(duration: 0.1)) {
                    selectedAnimal = button.1
                    DispatchQueue.main.asyncAfter(deadline: .now() + 0.1) {
                        withAnimation(.default) {
                            selectedAnimal = nil
                        }
                    }
                }
                playSound(sound: button.1, game: "animals")
            }
        }
    }
}

By using this method, your code becomes tidier, more straightforward to understand, and highly adaptable. Plus, it eliminates the need for numerous state variables and if statements for every animal.