http://www.amazon.com/gp/mas/dl/android?p=<your app package name>

Documentation: https://developer.amazon.com/docs/reports-promo/deeplink-to-the-amazon-client.html

The above answers did help me, however it gave me mixed results.

For some reason using the amazonToAlipay:// did not let me reach the desired product.

Using universal links, while they should work in theory, gave me mixed results. On some devices where the amazon app was installed, the amazon app opened without problems however on others, it opened in Safari even though the Amazon app was installed.

This could be because of device type, iOS versions or something else, go figure.

So as a work around, the following gave me the best results.

Using https://www.appsight.io/app/amazon as Milan suggested, I used the following scheme which was mentioned there:

com.amazon.mobile.shopping

Step 1. Whitelist this URL Scheme in your info.Plist by adding the following

<key>LSApplicationQueriesSchemes</key>
<array>
   <string>com.amazon.mobile.shopping</string>
</array>

Step 2. In my case, I needed to open a specific product on Amazon and get the product id which is present in the URL and build your string in the following format:

com.amazon.mobile.shopping://www.amazon.com/products/{your-product-id}/

Step 3. Try to open the amazon app url but also handle the case where the app might not be installed

func openAmazonProduct(withId id: String) {

   guard let amazonWebURL = URL(string: "https://amzn.to/2MQC8Bz"),
         let amazonAppURL = URL(string: "com.amazon.mobile.shopping://www.amazon.com/products/\(id)/") else {
             return
   }
        
   if UIApplication.shared.canOpenURL(amazonAppURL) {
            UIApplication.shared.open(amazonAppURL, options: [:], completionHandler: nil)
   }
   else if UIApplication.shared.canOpenURL(amazonWebURL) {
            UIApplication.shared.open(amazonWebURL, options: [:], completionHandler: nil)
   }

}